NCERT Solutions for Class 12 Maths Chapter 4 Determinants In Hindi pdf download
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Access NCERT Solutions for Mathematics Chapter 4 – सारणिक
प्रश्नावली 4.1
प्रश्न 1 से 2 तक मे सारणिकों का मान ज्ञात कीजिए:
1. \[\left| {\begin{array}{*{20}{c}} { {2}}&{ {4}} \\ {{ { - 5}}}&{{ { - 1}}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {2}}&{ {4}} \\ {{ { - 5}}}&{{ { - 1}}} \end{array}} \right|\] = \[{ {2( - 1) - 4( - 5) = - 2 + 20 = 18}}\]
2. (i) \[\left| {\begin{array}{*{20}{c}} {{ {cos\theta }}}&{{ { - sin\theta }}} \\ {{ {sin\theta }}}&{{ {cos\theta }}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} {{ {cos\theta }}}&{{ { - sin\theta }}} \\ {{ {sin\theta }}}&{{ {cos\theta }}} \end{array}} \right|\] = \[ (cos\theta \times cos\theta ) - (sin\theta \times ( - sin\theta ))\]
\[{ { = co}}{{ {s}}^{ {2}}}{ {\theta + si}}{{ {n}}^{ {2}}}{ {\theta = 1}}\]
(ii) \[\left| {\begin{array}{*{20}{c}} {{{ {x}}^{ {2}}}{ { - x + 1}}}&{{ {x - 1}}} \\ {{ {x + 1}}}&{{ {x + 1}}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} {{{ {x}}^{ {2}}}{ { - x + 1}}}&{{ {x - 1}}} \\ {{ {x + 1}}}&{{ {x + 1}}} \end{array}} \right|{ { = }}\left( {{{ {x}}^{ {2}}}{ { - x + 1}}} \right){ { \times (x + 1) - (x + 1) \times (x - 1)}}\]
\[\begin{align} { { = }}{{ {x}}^{ {3}}}{ { + }}{{ {x}}^{ {2}}}{ { - }}{{ {x}}^{ {2}}}{ { - x + x + 1 - }}\left( {{{ {x}}^{ {2}}}{ { + x - x - 1}}} \right) \hfill \\ { { = }}{{ {x}}^{ {3}}}{ { - }}{{ {x}}^{ {2}}}{ { + 2}} \hfill \\ \end{align} \]
3. यदि \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {2}} \\ { {4}}&{ {2}} \end{array}} \right]\] तो दिखाइए \[{ {|2A| = 4|A}}\mid \]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {2}} \\ { {4}}&{ {2}} \end{array}} \right]\]
\[{ {|2A| = }}\left| {\begin{array}{*{20}{l}} { {2}}&{ {4}} \\ { {8}}&{ {4}} \end{array}} \right|{ { = 2(4) - 8(4) = 8 - 32 = - 24}}\] .......(i)
\[{ {4|A| = 4}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {2}} \\ { {4}}&{ {2}} \end{array}} \right|{ { = 4[1(2) - 2(4)] = 4(2 - 8) = - 24}}\] .......(ii)
अतः (i) और (ii) से \[{ {|2A| = 4|A}}\mid \]
4. यदि \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {1}} \\ { {0}}&{ {1}}&{ {2}} \\ { {0}}&{ {0}}&{ {4}} \end{array}} \right]\] हो, तो दिखाइए \[{ {|3A| = 27|A}}\mid \]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {1}} \\ { {0}}&{ {1}}&{ {2}} \\ { {0}}&{ {0}}&{ {4}} \end{array}} \right]\]
\[{ {|3A| = }}\left| {\begin{array}{*{20}{c}} { {3}}&{ {0}}&{ {3}} \\ { {0}}&{ {3}}&{ {6}} \\ { {0}}&{ {0}}&{{ {12}}} \end{array}} \right|{ { = 3(36 - 0) - 0(0 - 0) + 3(0 - 0) = 108}}\] .......(i)
\[{ {27|A| = 27}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {1}} \\ { {0}}&{ {1}}&{ {2}} \\ { {0}}&{ {0}}&{ {4}} \end{array}} \right|{ { = 27[1(4 - 0) - 0(0 - 0) - 1(0 - 0)] = 27(4) = 108}}\] .......(ii)
अतः (i) और (ii) से \[{ {|3A| = 27|A}}\mid \]
5. निम्नलिखित सारणिकों का मान ज्ञात कीजिए:
(i) \[\left| {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}}&{{ { - 2}}} \\ { {0}}&{ {0}}&{{ { - 1}}} \\ { {3}}&{{ { - 5}}}&{ {0}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}}&{{ { - 2}}} \\ { {0}}&{ {0}}&{{ { - 1}}} \\ { {3}}&{{ { - 5}}}&{ {0}} \end{array}} \right|{ { = 3(0 - 5) + 1(0 + 3) + 2(0 - 0)}}\]
\[\begin{align} { { = - 15 + 3 - 0}} \hfill \\ { { = - 12}} \hfill \\ \end{align} \]
(ii) \[\left| {\begin{array}{*{20}{c}} { {3}}&{{ { - 4}}}&{ {5}} \\ { {1}}&{ {1}}&{{ { - 2}}} \\ { {2}}&{ {3}}&{ {1}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {3}}&{{ { - 4}}}&{ {5}} \\ { {1}}&{ {1}}&{{ { - 2}}} \\ { {2}}&{ {3}}&{ {1}} \end{array}} \right|{ { = 3(1 + 6) + 4(1 + 4) + 5(3 - 2)}}\]
\[\begin{align} { { = 21 + 20 + 5}} \hfill \\ { { = 46}} \hfill \\ \end{align} \]
(iii) \[\left| {\begin{array}{*{20}{c}} { {0}}&{ {1}}&{ {2}} \\ {{ { - 1}}}&{ {0}}&{{ { - 3}}} \\ {{ { - 2}}}&{ {3}}&{ {0}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {0}}&{ {1}}&{ {2}} \\ {{ { - 1}}}&{ {0}}&{{ { - 3}}} \\ {{ { - 2}}}&{ {3}}&{ {0}} \end{array}} \right|\] = \[{ {0(0 + 9) - 1(0 - 6) + 2( - 3 - 0)}}\]
\[{ { = 0 + 6 - 6 = 0}}\]
(iv) \[\left| {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{{ { - 2}}} \\ { {0}}&{ {2}}&{{ { - 1}}} \\ { {3}}&{{ { - 5}}}&{ {0}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{{ { - 2}}} \\ { {0}}&{ {2}}&{{ { - 1}}} \\ { {3}}&{{ { - 5}}}&{ {0}} \end{array}} \right|{ { = 2(0 - 5) + 1(0 + 3) - 2(0 - 6)}}\]
\[\begin{align} { { = - 10 + 3 + 12}} \hfill \\ { { = 5}} \hfill \\ \end{align} \]
6. यदि \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {1}}&{ {2}} \\ { {2}}&{ {1}}&{ {3}} \\ { {5}}&{ {4}}&{ {9}} \end{array}} \right]\] हो, तो ज्ञात कीजिए \[{ {|A|}}\]
उत्तर: \[{ {|A| = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {1}}&{ {2}} \\ { {2}}&{ {1}}&{ {3}} \\ { {5}}&{ {4}}&{ {9}} \end{array}} \right|\]
\[\begin{align} { { = 1( - 9 + 12) - 1( - 18 + 15) - 2(8 - 5)}} \hfill \\ { { = 3 + 3 - 6}} \hfill \\ { { = 0}} \hfill \\ \end{align} \]
7. \[{ {x}}\] के मान ज्ञात कीजिए यदि
(i) \[\left| {\begin{array}{*{20}{l}} { {2}}&{ {4}} \\ { {5}}&{ {1}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{c}} {{ {2x}}}&{ {4}} \\ { {6}}&{ {x}} \end{array}} \right|\]
उत्तर: \[{ {2(1) - 5(4) = 2x(x) - 6(4)}}\]
\[\begin{align} { {2 - 20 = 2}}{{ {x}}^{ {2}}}{ { - 24}} \hfill \\ { { - 18 = 2}}{{ {x}}^{ {2}}}{ { - 24}} \hfill \\ { {2}}{{ {x}}^{ {2}}}{ { - 24 + 18 = 0}} \hfill \\ \end{align} \]
\[{ {2}}{{ {x}}^{ {2}}}{ { - 6 = 0}}\]
\[\begin{align} {{ {x}}^{ {2}}}{ { = }}\dfrac{{ {6}}}{{ {2}}}{ { = 3}} \hfill \\ { {x = \pm }}\sqrt { {3}} \hfill \\ \end{align} \]
(ii) \[\left| {\begin{array}{*{20}{l}} { {2}}&{ {3}} \\ { {4}}&{ {5}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{c}} { {x}}&{ {3}} \\ {{ {2x}}}&{ {5}} \end{array}} \right|\]
उत्तर: \[{ {2(5) - 4(3) = x(5) - 3(2x)}}\]
\[\begin{align} { {10 - 12 = 5x - 6x}} \hfill \\ { { - 2 = - x}} \hfill \\ { {x = 2}} \hfill \\ \end{align} \]
8. यदि \[\left| {\begin{array}{*{20}{c}} { {x}}&{ {2}} \\ {{ {18}}}&{ {x}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{c}} { {6}}&{ {2}} \\ {{ {18}}}&{ {6}} \end{array}} \right|\] हो तो \[{ {x}}\] बराबर है
(a) \[6\]
(b) \[ \pm 6\]
(c) \[ - 6\]
(d) \[0\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {x}}&{ {2}} \\ {{ {18}}}&{ {x}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{c}} { {6}}&{ {2}} \\ {{ {18}}}&{ {6}} \end{array}} \right|\]
\[\begin{align} {{ {x}}^{ {2}}}{ { - 36 = 36 - 36}} \hfill \\ {{ {x}}^{ {2}}}{ { = 36}} \hfill \\ { {x = \pm 6}} \hfill \\ \end{align} \]
इसलिए सही विकल्प (b) है।
प्रश्नावली 4.2
बिना प्रसारण किए और सारणिकों के गुणधर्मों का प्रयोग करके निम्नलिखित प्रश्न 1 से 5 को सिद्ध कीजिए।
1. \[\left| {\begin{array}{*{20}{l}} { {x}}&{ {a}}&{{ {x + a}}} \\ { {y}}&{ {b}}&{{ {y + b}}} \\ { {z}}&{ {c}}&{{ {z + c}}} \end{array}} \right|{ { = 0}}\]
उत्तर: \[\left| {\begin{array}{*{20}{l}} { {x}}&{ {a}}&{{ {x + a}}} \\ { {y}}&{ {b}}&{{ {y + b}}} \\ { {z}}&{ {c}}&{{ {z + c}}} \end{array}} \right|\] = \[\left| {\begin{array}{*{20}{l}} {{ {x + a}}}&{ {a}}&{{ {x + a}}} \\ {{ {y + b}}}&{ {b}}&{{ {y + b}}} \\ {{ {z + c}}}&{ {c}}&{{ {z + c}}} \end{array}} \right|\;\;\;\,\,\,\left[ {\because \;{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}} \right]\]
\[{ { = 0}}\;{ { }}\left[ {\because \;{{ {C}}_{ {1}}}{ { = }}{{ {C}}_{ {3}}}} \right]\]
2. \[\left| {\begin{array}{*{20}{l}} {{ {a - b}}}&{{ {b - c}}}&{{ {c - a}}} \\ {{ {b - c}}}&{{ {c - a}}}&{{ {a - b}}} \\ {{ {c - a}}}&{{ {a - b}}}&{{ {b - c}}} \end{array}} \right|{ { = 0}}\]
उत्तर: \[{ { = }}\left| {\begin{array}{*{20}{l}} { {0}}&{{ {b - c}}}&{{ {c - a}}} \\ { {0}}&{{ {c - a}}}&{{ {a - b}}} \\ { {0}}&{{ {a - b}}}&{{ {b - c}}} \end{array}} \right|\;\;\;\;\;\;\;\;\,\left[ {{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}{ { + }}{{ {C}}_{ {3}}}} \right]\]
\[{ { = 0}}\;\;\;\,{ { }}\,\,\;{ { }}\left[ {{{ {C}}_{ {1}}}{ { = 0}}} \right]\]
3. \[\left| {\begin{array}{*{20}{l}} { {2}}&{ {7}}&{{ {65}}} \\ { {3}}&{ {8}}&{{ {75}}} \\ { {5}}&{ {9}}&{{ {86}}} \end{array}} \right|{ { = 0}}\]
उत्तर: \[\left| {\begin{array}{*{20}{l}} { {2}}&{ {7}}&{{ {65}}} \\ { {3}}&{ {8}}&{{ {75}}} \\ { {5}}&{ {9}}&{{ {86}}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{l}} { {2}}&{ {7}}&{{ {63}}} \\ { {3}}&{ {8}}&{{ {72}}} \\ { {5}}&{ {9}}&{{ {81}}} \end{array}} \right|\;\;\;\;\;\,\;\;\;\;\;\left[ {{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}}} \right]\]
\[{ { = 9}}\left| {\begin{array}{*{20}{l}} { {2}}&{ {7}}&{ {7}} \\ { {3}}&{ {8}}&{ {8}} \\ { {5}}&{ {9}}&{ {9}} \end{array}} \right|\] \[{{C}_{3}}\] से \[9\] उभयनिष्ट लेने पर]
\[{ { = 0}}\;\,{ { }}\;\left[ {{{ {C}}_{ {2}}}{ { = }}{{ {C}}_{ {3}}}} \right]\]
4. \[\left| {\begin{array}{*{20}{l}} { {1}}&{{ {bc}}}&{{ {a(b + c)}}} \\ { {1}}&{{ {ca}}}&{{ {b(c + a)}}} \\ { {1}}&{{ {ab}}}&{{ {c(a + b)}}} \end{array}} \right|{ { = 0}}\]
उत्तर: \[\left| {\begin{array}{*{20}{l}} { {1}}&{{ {bc}}}&{{ {a(b + c)}}} \\ { {1}}&{{ {ca}}}&{{ {b(c + a)}}} \\ { {1}}&{{ {ab}}}&{{ {c(a + b)}}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{{ {bc}}}&{{ {ab + bc + ca}}} \\ { {1}}&{{ {ca}}}&{{ {ab + bc + ca}}} \\ { {1}}&{{ {ab}}}&{{ {ab + bc + ca}}} \end{array}} \right|\] \[\left[ {{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { + }}{{ {C}}_{ {2}}}} \right]\]
\[{ { = (ab + bc + ca)}}\left| {\begin{array}{*{20}{l}} { {1}}&{{ {bc}}}&{ {1}} \\ { {1}}&{{ {ca}}}&{ {1}} \\ { {1}}&{{ {ab}}}&{ {1}} \end{array}} \right|\]
(\[{{ {C}}_{ {3}}}\] से \[\left( {{ {ab + bc + }}} \right.{ {ca)}}\] उभयनिष्ठ लेने पर)
\[{ { = 0}}\;\,{ { }}\;{ { }}\;{ { }}\left[ {{{ {C}}_{ {1}}}{ { = }}{{ {C}}_{ {3}}}} \right]\]
5. \[\left| {\begin{array}{*{20}{l}} {{ {b + c}}}&{{ {q + r}}}&{{ {y + z}}} \\ {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\ {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}} \end{array}} \right|{ { = 2}}\left| {\begin{array}{*{20}{l}} { {a}}&{ {p}}&{ {x}} \\ { {b}}&{ {q}}&{ {y}} \\ { {c}}&{ {r}}&{ {z}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{l}} {{ {b + c}}}&{{ {q + r}}}&{{ {y + z}}} \\ {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\ {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}} \end{array}} \right|{ { = 2}}\left| {\begin{array}{*{20}{l}} { {a}}&{ {p}}&{ {x}} \\ { {b}}&{ {q}}&{ {y}} \\ { {c}}&{ {r}}&{ {z}} \end{array}} \right|\]
LHS \[{ { = }}\left| {\begin{array}{*{20}{l}} {{ {b + c}}}&{{ {q + r}}}&{{ {y + z}}} \\ {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\ {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {2c}}}&{{ {2r}}}&{{ {2z}}} \\ {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\ {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}} \end{array}} \right|\;\;\;\,\,\,\;\;\;\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right]\]
\[{ { = 2}}\left| {\begin{array}{*{20}{c}} { {c}}&{ {r}}&{ {z}} \\ {{ {c + a}}}&{{ {r + p}}}&{{ {z + x}}} \\ {{ {a + b}}}&{{ {p + q}}}&{{ {x + y}}} \end{array}} \right|\]
\[{{R}_{1}}\] से \[ 2 \] उभयनिष्ट लेने पर]
\[{ { = 2}}\left| {\begin{array}{*{20}{c}} { {c}}&{ {r}}&{ {z}} \\ { {a}}&{ {p}}&{ {x}} \\ {{ {a + b}}}&{{ {p + q}}\quad }&{{ {x + y}}} \end{array}} \right|\;\;\;\;\;\;\left[ {{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {1}}}} \right]\]
\[{ { = 2}}\left| {\begin{array}{*{20}{l}} { {c}}&{ {r}}&{ {z}} \\ { {a}}&{ {p}}&{ {x}} \\ { {b}}&{ {q}}&{ {y}} \end{array}} \right|\;\;\;\;\;\;\;\,\;\;\left[ {{{ {R}}_{ {3}}} \to {{ {R}}_{ {3}}}{ { - }}{{ {R}}_{ {2}}}} \right]\]
\[{ { = 2}}\left| {\begin{array}{*{20}{l}} { {a}}&{ {p}}&{ {x}} \\ { {c}}&{ {r}}&{ {z}} \\ { {b}}&{ {q}}&{ {y}} \end{array}} \right|\;\;\,\;\,\,\;\;\;\,\;\left[ {{{ {R}}_{ {1}}} \leftrightarrow {{ {R}}_{ {2}}}} \right]\]
\[{ { = 2}}\left| {\begin{array}{*{20}{l}} { {a}}&{ {p}}&{ {x}} \\ { {b}}&{ {q}}&{ {z}} \\ { {c}}&{ {r}}&{ {y}} \end{array}} \right|\;\;\,\;\,\,\;\;\;\,\;\left[ {{{ {R}}_2} \leftrightarrow {{ {R}}_3}} \right]\]
= RHS
सारणिकों के गुणधर्मों का प्रयोग करके प्रश्न 6 से 14 तक को सिद्ध कीजिए।
6. \[\left| {\begin{array}{*{20}{c}} { {0}}&{ {a}}&{{ { - b}}} \\ {{ { - a}}}&{ {0}}&{{ { - c}}} \\ { {b}}&{ {c}}&{ {0}} \end{array}} \right|{ { = 0}}\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} { {0}}&{ {a}}&{{ { - b}}} \\ {{ { - a}}}&{ {0}}&{{ { - c}}} \\ { {b}}&{ {c}}&{ {0}} \end{array}} \right|\]
LHS = \[\left| {\begin{array}{*{20}{c}} { {0}}&{ {a}}&{{ { - b}}} \\ {{ { - a}}}&{ {0}}&{{ { - c}}} \\ { {b}}&{ {c}}&{ {0}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} { {0}}&{ {a}}&{{ { - b}}} \\ {{ { - ab}}}&{ {0}}&{{ { - bc}}} \\ {{ {ab}}}&{{ {ac}}}&{ {0}} \end{array}} \right|\;\,\,\;\;\;\;\,\;\;\;\left[ {{{ {R}}_{ {2}}} \to { {b}}{{ {R}}_{ {2}}}{ { , }}{{ {R}}_{ {3}}} \to { {a}}{{ {R}}_{ {3}}}} \right] \]
\[{ { = }}\left| {\begin{array}{*{20}{c}} { {0}}&{ {a}}&{{ { - b}}} \\ { {0}}&{{ {ac}}}&{{ { - bc}}} \\ {{ {ab}}}&{{ {ac}}}&{ {0}} \end{array}} \right|\;\;\;\;\;\;\;\;\,\;\;\left[ {{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { + 3}}} \right]\]
\[= ab( - abc + abc) \]
\[{ { = ab(0)}} \]
\[{ { = 0}}\]
= RHS
7. \[\left| {\begin{array}{*{20}{c}} {{ { - }}{{ {a}}^{ {2}}}}&{{ {ab}}}&{{ {ac}}} \\ {{ {ba}}}&{{ { - }}{{ {b}}^{ {2}}}}&{{ {bc}}} \\ {{ {ca}}}&{{ {cb}}}&{{ { - }}{{ {c}}^{ {2}}}} \end{array}} \right|{ { = 4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ { - }}{{ {a}}^{ {2}}}}&{{ {ab}}}&{{ {ac}}} \\ {{ {ba}}}&{{ { - }}{{ {b}}^{ {2}}}}&{{ {bc}}} \\ {{ {ca}}}&{{ {cb}}}&{{ { - }}{{ {c}}^{ {2}}}} \end{array}} \right|\]
\[{ { = abc}}\left| {\begin{array}{*{20}{c}} {{ { - a}}}&{ {a}}&{ {a}} \\ { {b}}&{ {-b}}&{ {b}} \\ { {c}}&{ {c}}&{{ { - c}}} \end{array}} \right|\]
\[{{ {C}}_{ {1}}}{ {,}}\;{{ {C}}_{ {2}}}{ {,}}\;{{ {C}}_{ {3}}}\] से \[{ {a,}}\;{ {b,}}\;{ {c}}\] उभयनिष्ट लेने पर]
\[{ { = }}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\left| {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {1}}&{ {1}} \\ { {1}}&{ {-1}}&{ {1}} \\ { {1}}&{ {1}}&{{ { - 1}}} \end{array}} \right|\] \[{{ {R}}_{ {1}}}{ {,}}\;{{ {R}}_{ {2}}}{ {,}}\;{{ {R}}_{ {3}}}\] से \[{ {a,}}\;{ {b,}}\;{ {c}}\] उभयनिष्ट लेने पर]
\[{ { = }}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\left| {\begin{array}{*{20}{c}} { {0}}&{ {1}}&{ {1}} \\ { {0}}&{{ { - 1}}}&{ {1}} \\ { {2}}&{ {1}}&{{ { - 1}}} \end{array}} \right|\;\;\;\;\;\;\,\;\;\;\;\;\left[ {{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}} \right]\]
\[{ { = }}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}{ {\{ 2(1 + 1)\} }}\] \[{{ {C}}_1}\] के अनुदिश प्रसरण करने पर]
\[{ { = 4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\]
= RHS
8. (i) \[\left| {\begin{array}{*{20}{l}} { {1}}&{ {a}}&{{{ {a}}^{ {2}}}} \\ { {1}}&{ {b}}&{{{ {b}}^{ {2}}}} \\ { {1}}&{ {c}}&{{{ {c}}^{ {2}}}} \end{array}} \right|{ { = (a - b)(b - c)(c - a)}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}} { {1}}&{ {a}}&{{{ {a}}^{ {2}}}} \\ { {1}}&{ {b}}&{{{ {b}}^{ {2}}}} \\ { {1}}&{ {c}}&{{{ {c}}^{ {2}}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} { {0}}&{{ {a - b}}}&{{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}} \\ { {0}}&{{ {b - c}}}&{{{ {b}}^{ {2}}}{ { - }}{{ {c}}^{ {2}}}} \\ { {1}}&{ {c}}&{{{ {c}}^{ {2}}}} \end{array}} \right|\;\,\;\;\;\;\,\;\;\,\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right]\]
\[{ { = (a - b)(b - c)}}\left| {\begin{array}{*{20}{c}} { {0}}&{ {1}}&{{ {a + b}}} \\ { {0}}&{ {1}}&{{ {b + c}}} \\ { {1}}&{ {c}}&{{{ {c}}^{ {2}}}} \end{array}} \right|\] \[{{ {R}}_{ {1}}}\] से \[{ {a - b, }}{{ {R}}_{ {2}}}\] से \[{ {b - c}}\] उभयनिष्ट लेने पर]
\[{ { = (a - b)(b - c)\{ 1(b + c - a - b)\} }}\] \[{{ {C}}_{ {1}}}\] के अनुदिश प्रसरण करने पर]
\[{ { = (a - b)(b - c)(c - a)}}\]
(ii) \[\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {a}}&{ {b}}&{ {c}} \\ {{{ {a}}^{ {3}}}}&{{{ {b}}^{ {3}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|{ { = (a - b)(b - c)(c - a)(a + b + c)}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {a}}&{ {b}}&{ {c}} \\ {{{ {a}}^{ {3}}}}&{{{ {b}}^{ {3}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} { {0}}&{ {0}}&{ {1}} \\ {{ {a - b}}}&{{ {b - c}}}&{ {c}} \\ {{{ {a}}^{ {3}}}{ { - }}{{ {b}}^{ {3}}}}&{{{ {b}}^{ {3}}}{ { - }}{{ {c}}^{ {3}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|\;\;\;\;\;\,\;\;\left[ {{{ {C}}_{ {1}}} \to } \right.\left. {{{ {C}}_{ {1}}}{ { - }}{{ {C}}_{ {2}}}{ {, }}{{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {3}}}} \right]\]
\[{ { = (a - b)(b - c)}}\left| {\begin{array}{*{20}{c}} { {0}}&{ {0}}&{ {1}} \\ { {1}}&{ {1}}&{ {c}} \\ {{{ {a}}^{ {2}}}{ { + ab + }}{{ {b}}^{ {2}}}}&{{{ {b}}^{ {2}}}{ { + bc + }}{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|\] \[{{ {C}}_{ {1}}}\] से \[{ {a - b}}\] , \[{{ {C}}_{ {2}}}\] से \[{ {b - c}}\] उभयनिष्ट लेने पर]
\[{ { = (a - b)(b - c)}}\left\{ {{ {1}}\left( {{{ {b}}^{ {2}}}{ { + bc + }}{{ {c}}^{ {2}}}} \right){ { - }}\left( {{{ {a}}^{ {2}}}{ { + ab + }}{{ {b}}^{ {2}}}} \right)} \right\}\] \[{{ {R}}_{ {1}}}\] के अनुदिश प्रसरण करने पर]
\[{ { = (a - b)(b - c)(c - a)(a + b + c)}}\]
= RHS
9. \[\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{{ {yz}}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{{ {zx}}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{{ {xy}}} \end{array}} \right|{ { = (x - y)(y - z)(z - x)(xy + yz + zx)}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{{ {yz}}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{{ {zx}}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{{ {xy}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{l}} {{{ {x}}^{ {2}}}}&{{{ {x}}^{ {3}}}}&{{ {xyz}}} \\ {{{ {y}}^{ {2}}}}&{{{ {y}}^{ {3}}}}&{{ {yzx}}} \\ {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {3}}}}&{{ {zxy}}} \end{array}} \right|\;\;\;\,\,\,\,\,\;\;\;\left[ {{{ {R}}_{ {1}}} \to { {x}}{{ {R}}_{ {1}}},\;} \right.\left. {{{ {R}}_{ {2}}} \to { {y}}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {3}}} \to { {y}}{{ {R}}_{ {3}}}} \right]\]
\[{ { = xyz}}\left| {\begin{array}{*{20}{l}} {{{ {x}}^{ {2}}}}&{{{ {x}}^{ {3}}}}&{ {1}} \\ {{{ {y}}^{ {2}}}}&{{{ {y}}^{ {3}}}}&{ {1}} \\ {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {3}}}}&{ {1}} \end{array}} \right|\] \[{{ {C}}_{ {3}}}\] से \[{ {xyz}}\] के उभयनिष्ट लेने पर]
\[{ { = xyz}}\left| {\begin{array}{*{20}{c}} {{{ {x}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}}&{{{ {x}}^{ {3}}}{ { - }}{{ {y}}^{ {3}}}}&{ {0}} \\ {{{ {y}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}}&{{{ {y}}^{ {3}}}{ { - }}{{ {z}}^{ {3}}}}&{ {0}} \\ {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {2}}}}&{ {1}} \end{array}} \right|\;\;\;\,\;\;\;\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right] \]
\[{ { = xyz(x - y)(y - z)}}\left| {\begin{array}{*{20}{c}} {{ {x + y}}}&{{{ {x}}^{ {2}}}{ { + xy + }}{{ {y}}^{ {2}}}}&{ {0}} \\ {{ {y + z}}}&{{{ {y}}^{ {2}}}{ { + yz + }}{{ {z}}^{ {2}}}}&{ {0}} \\ {{{ {z}}^{ {2}}}}&{{{ {z}}^{ {2}}}}&{ {1}} \end{array}} \right|\;\;\;\;\;\left[ {{{ {R}}_{ {1}}} \to { {x - y, }}{{ {R}}_{ {2}}} \to { {y - z}}} \right] \]
\[{ { = xyz(x - y)(y - z)\{ (x + y)(}}{{ {y}}^{ {2}}}{ { + yz + }}{{ {z}}^{ {2}}}{ {) - (y + z)(}}{{ {x}}^{ {2}}}{ { + xy + }}{{ {y}}^{ {2}}}{ {)\} }}\] \[{{ {C}}_{ {3}}}\] के अनुदिश प्रसरण करने पर]
\[{ { = (x - y)(y - z)(z - x)(xy + yz + zx)}}\]
= RHS
10. (i) \[\left| {\begin{array}{*{20}{c}} {{ {x + 4}}}&{{ {2x}}}&{{ {2x}}} \\ {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\ {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}} \end{array}} \right|{ { = (5x + 4)(4 - x}}{{ {)}}^{ {2}}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {x + 4}}}&{{ {2x}}}&{{ {2x}}} \\ {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\ {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}} \end{array}} \right|\]
\[{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \]
= \[\left| {\begin{array}{*{20}{c}} {{ {5x + 4}}}&{{ {5x + 4}}}&{{ {5x + 4}}} \\ {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\ {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}} \end{array}} \right| \]
\[{ {(5x + 4)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ {{ {2x}}}&{{ {x + 4}}}&{{ {2x}}} \\ {{ {2x}}}&{{ {2x}}}&{{ {x + 4}}} \end{array}} \right|\] \[{{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}}\]
\[{ { = }}\;{ {(5x + 4)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{ {2x}}}&{{ { - x + 4}}}&{ {0}} \\ {{ {2x}}}&{ {0}}&{{ { - x + 4}}} \end{array}} \right|\]
\[{ { = }}\;{ {(5x + 4)(4 - x)(4 - x)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{ {2x}}}&{ {1}}&{ {0}} \\ {{ {2x}}}&{ {0}}&{ {1}} \end{array}} \right|\]
\[{ { = }}\;{ {(5x + 4)(4 - x}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}} \\ {{ {2x}}}&{ {1}} \end{array}} \right| \]
\[ { { = }}\;{ {(5x + 4)(4 - x}}{{ {)}}^{ {2}}}\]
= RHS
(ii) \[\left| {\begin{array}{*{20}{c}} {{ {y + k}}}&{ {y}}&{ {y}} \\ { {y}}&{{ {y + k}}}&{ {y}} \\ { {y}}&{ {y}}&{{ {y + k}}} \end{array}} \right|{ { = }}{{ {k}}^{ {2}}}{ {(3y + k)}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {y + k}}}&{ {y}}&{ {y}} \\ { {y}}&{{ {y + k}}}&{ {y}} \\ { {y}}&{ {y}}&{{ {y + k}}} \end{array}} \right|\]
\[\begin{align} {{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \hfill \\ = \;\left| {\begin{array}{*{20}{c}} {{ {3y + k}}}&{{ {3y + k}}}&{{ {3y + k}}} \\ { {y}}&{{ {y + k}}}&{ {y}} \\ { {y}}&{ {y}}&{{ {y + k}}} \end{array}} \right| \hfill \\ \end{align} \]
\[{ { = }}\;{ {(3y + k)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {y}}&{{ {y + k}}}&{ {y}} \\ { {y}}&{ {y}}&{{ {y + k}}} \end{array}} \right|\]
\[\begin{align} {{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}} \hfill \\ { { = }}\;{ {(3y + k)}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {0}} \\ { {y}}&{ {k}}&{ {0}} \\ { {y}}&{ {0}}&{ {k}} \end{array}} \right| \hfill \\ \end{align} \]
\[\begin{align} { { = }}\;{ {(3y + k)}}{{ {k}}^{ {2}}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {0}} \\ { {y}}&{ {1}}&{ {0}} \\ { {y}}&{ {0}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\;{ {(3y + k)}}{{ {k}}^{ {2}}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {y}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\;{ {(3y + k)}}{{ {k}}^{ {2}}} \hfill \\ \end{align} \]
= RHS
11. (i) \[\left| {\begin{array}{*{20}{c}} {{ {a - b - c}}}&{{ {2a}}}&{{ {2a}}} \\ {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\ {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}} \end{array}} \right|{ { = (a + b + c}}{{ {)}}^3}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {a - b - c}}}&{{ {2a}}}&{{ {2a}}} \\ {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\ {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}} \end{array}} \right|\]
\[\begin{align} {{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \hfill \\ = \;\left| {\begin{array}{*{20}{c}} {{ {a + b + c}}}&{{ {a + b + c}}}&{{ {a + b + c}}} \\ {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\ {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}} \end{array}} \right| \hfill \\ { { = }}\;{ {(a + b + c)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ {{ {2b}}}&{{ {b - c - a}}}&{{ {2b}}} \\ {{ {2c}}}&{{ {2c}}}&{{ {c - a - b}}} \end{array}} \right| \hfill \\ {{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}} \hfill \\ \end{align} \]
\[\begin{align} { { = }}\;{ {(a + b + c)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{ {2b}}}&{{ { - (a + b + c)}}}&{ {0}} \\ {{ {2c}}}&{ {0}}&{{ { - (a + b + c)}}} \end{array}} \right| \hfill \\ { { = }}\;{{ {(a + b + c)}}^{ {3}}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{ {2b}}}&{{ { - 1}}}&{ {0}} \\ {{ {2c}}}&{ {0}}&{{ { - 1}}} \end{array}} \right| \hfill \\ { { = }}\;{{ {(a + b + c)}}^{ {3}}}{ {( - 1) \times ( - 1)}}\;{ { = (a + b + c}}{{ {)}}^{ {3}}} \hfill \\ \end{align} \]
= RHS
(ii) \[\left| {\begin{array}{*{20}{c}} {{ {x + y + 2z}}}&{ {x}}&{ {y}} \\ { {z}}&{{ {y + z + 2x}}}&{ {y}} \\ { {z}}&{ {x}}&{{ {z + x + 2y}}} \end{array}} \right|{ { = 2(x + y + z}}{{ {)}}^{ {3}}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {x + y + 2z}}}&{ {x}}&{ {y}} \\ { {z}}&{{ {y + z + 2x}}}&{ {y}} \\ { {z}}&{ {x}}&{{ {z + x + 2y}}} \end{array}} \right|\]
\[\begin{align} {{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}{ { + }}{{ {C}}_{ {3}}} \hfill \\ = \;\left| {\begin{array}{*{20}{c}} {{ {2(x + y + z)}}}&{ {x}}&{ {y}} \\ {{ {2(x + y + z)}}}&{{ {y + z + 2x}}}&{ {y}} \\ {{ {2(x + y + z)}}}&{ {x}}&{{ {z + x + 2y}}} \end{array}} \right| \hfill \\ { { = }}\;{ {2(x + y + z)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {x}}&{ {y}} \\ { {1}}&{{ {y + z + 2x}}}&{ {y}} \\ { {1}}&{ {x}}&{{ {z + x + 2y}}} \end{array}} \right| \hfill \\ {{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {1}}}{ {, }}{{ {R}}_{ {3}}} \to {{ {R}}_{ {3}}}{ { - }}{{ {R}}_{ {1}}} \hfill \\ \end{align} \]
\[\begin{align} { { = }}\;{ {2(x + y + z)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {x}}&{ {y}} \\ { {0}}&{{ {y + z + x}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {z + x + y}}} \end{array}} \right| \hfill \\ { { = 2(x + y + z}}{{ {)}}^{ {3}}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {x}}&{ {y}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right| \hfill \\ { { = 2(x + y + z}}{{ {)}}^{ {3}}} \hfill \\ \end{align} \]
= RHS
12. \[\left| {\begin{array}{*{20}{c}} { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\ {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\ { {x}}&{{{ {x}}^{ {2}}}}&{ {1}} \end{array}} \right|{ { = }}{\left( {{ {1 + }}{{ {x}}^{ {3}}}} \right)^{ {2}}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\ {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\ { {x}}&{{{ {x}}^{ {2}}}}&{ {1}} \end{array}} \right|\]
\[\begin{align} {{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {2}}}{ { + }}{{ {R}}_{ {3}}} \hfill \\ = \;\left| {\begin{array}{*{20}{c}} {{ {1 + x + }}{{ {x}}^{ {2}}}}&{{ {1 + x + }}{{ {x}}^{ {2}}}}&{{ {1 + x + }}{{ {x}}^{ {2}}}} \\ {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\ { {x}}&{{{ {x}}^{ {2}}}}&{ {1}} \end{array}} \right| \hfill \\ \end{align} \]
\[\begin{align} = \;\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right)\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ {{{ {x}}^{ {2}}}}&{ {1}}&{ {x}} \\ { {x}}&{{{ {x}}^{ {2}}}}&{ {1}} \end{array}} \right| \hfill \\ {{ {C}}_{ {2}}} \to {{ {C}}_{ {2}}}{ { - }}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {3}}} \to {{ {C}}_{ {3}}}{ { - }}{{ {C}}_{ {1}}} \hfill \\ \end{align} \]
\[\begin{align} = \;\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right)\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{{ {x}}^{ {2}}}}&{{ {1 - }}{{ {x}}^{ {2}}}}&{{ {x - }}{{ {x}}^{ {2}}}} \\ { {x}}&{{{ {x}}^{ {2}}}{ { - x}}}&{{ {1 - }}{{ {x}}^{ {2}}}} \end{array}} \right| \hfill \\ = \,\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right){ {(1 - x)(1 - x)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{{ {x}}^{ {2}}}}&{{ {1 + x}}}&{ {x}} \\ { {x}}&{{ { - x}}}&{ {1}} \end{array}} \right| \hfill \\ = \,\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right){ {(1 - x)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{{ {x}}^{ {2}}}}&{{ {1 + x}}}&{ {x}} \\ { {x}}&{{ { - x}}}&{ {1}} \end{array}} \right| \hfill \\ = \;\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right){ {(1 - x)(1)}}\left| {\begin{array}{*{20}{c}} {{ {1 + x}}}&{ {x}} \\ {{ { - x}}}&{ {1}} \end{array}} \right| \hfill \\ = \;\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right){ {(1 - x)}}\left( {{ {1 + x + }}{{ {x}}^{ {2}}}} \right) \hfill \\ = \;{\left( {{ {1 - }}{{ {x}}^{ {3}}}} \right)^{ {2}}} \hfill \\ \end{align} \]
= RHS
13. \[\left| {\begin{array}{*{20}{c}} {{ {1 + }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}&{{ {2ab}}}&{{ { - 2b}}} \\ {{ {2ab}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{{ {2a}}} \\ {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}} \end{array}} \right|{ { = }}{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {3}}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {1 + }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}}&{{ {2ab}}}&{{ { - 2b}}} \\ {{ {2ab}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{{ {2a}}} \\ {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}} \end{array}} \right|\]
\[{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - a}}{{ {R}}_{ {3}}}{ {, }}{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { + b}}{{ {R}}_{ {3}}}\]
\[ = \;\left| {\begin{array}{*{20}{c}} {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{ {0}}&{{ { - b}}\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)} \\ { {0}}&{{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}}&{{ {a}}\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)} \\ {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}} \end{array}} \right|\]
\[ = \;{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {2}}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{{ { - b}}} \\ { {0}}&{ {1}}&{ {a}} \\ {{ {2b}}}&{{ { - 2a}}}&{{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}} \end{array}} \right|\]
\[ = \;{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {2}}}\left( {{ {1 - }}{{ {a}}^{ {2}}}{ { - }}{{ {b}}^{ {2}}}{ { + 2}}{{ {a}}^{ {2}}}{ { - b( - 2b)}}} \right)\]
\[\begin{align} = \,{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {2}}}\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right) \hfill \\ = \,{\left( {{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}} \right)^{ {3}}} \hfill \\ \end{align} \]
= RHS
14. \[\left| {\begin{array}{*{20}{c}} {{{ {a}}^{ {2}}}{ { + 1}}}&{{ {ab}}}&{{ {ac}}} \\ {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + 1}}}&{{ {bc}}} \\ {{ {ca}}}&{{ {cb}}}&{{{ {c}}^{ {2}}}{ { + 1}}} \end{array}} \right|{ { = 1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + }}{{ {c}}^{ {2}}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {1 + }}{{ {a}}^{ {2}}}}&{{ {ab}}}&{{ {ac}}} \\ {{ {ab}}}&{{ {1 + }}{{ {b}}^{ {2}}}}&{{ {bc}}} \\ {{ {ca}}}&{{ {cb}}}&{{{ {c}}^{ {2}}}{ { + 1}}} \end{array}} \right|\]
\[\begin{align} { { = }}\;{ {abc}}\left| {\begin{array}{*{20}{c}} {{ {a + }}\dfrac{{ {1}}}{{ {a}}}}&{ {b}}&{ {c}} \\ { {a}}&{{ {b + }}\dfrac{{ {1}}}{{ {b}}}}&{ {c}} \\ { {a}}&{ {b}}&{{ {c + }}\dfrac{{ {1}}}{{ {c}}}} \end{array}} \right| \hfill \\ {{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {1}}}{ {, }}{{ {R}}_{ {3}}} \to {{ {R}}_{ {3}}}{ { - }}{{ {R}}_{ {1}}} \hfill \\ \end{align} \]
\[\begin{align} { { = }}\;{ {abc}}\left| {\begin{array}{*{20}{c}} {{ {a + }}\dfrac{{ {1}}}{{ {a}}}}&{ {b}}&{ {c}} \\ {{ { - }}\dfrac{{ {1}}}{{ {a}}}}&{\dfrac{{ {1}}}{{ {b}}}}&{ {0}} \\ {{ { - }}\dfrac{{ {1}}}{{ {a}}}}&{ {0}}&{\dfrac{{ {1}}}{{ {c}}}} \end{array}} \right| \hfill \\ {{ {C}}_{ {1}}} \to { {a}}{{ {C}}_{ {1}}}{ {, }}{{ {C}}_{ {2}}} \to { {b}}{{ {C}}_{ {2}}}{ {, }}{{ {C}}_{ {3}}} \to { {c}}{{ {C}}_{ {3}}} \hfill \\ \end{align} \]
\[\begin{align} { { = }}\;{ {abc \times }}\dfrac{{ {1}}}{{{ {abc}}}}\left| {\begin{array}{*{20}{c}} {{ {1 + }}{{ {a}}^{ {2}}}}&{{{ {b}}^{ {2}}}}&{{{ {c}}^{ {2}}}} \\ {{ { - 1}}}&{ {1}}&{ {0}} \\ {{ { - 1}}}&{ {0}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\;{ { - 1}}\left| {\begin{array}{*{20}{c}} {{{ {b}}^{ {2}}}}&{{{ {c}}^{ {2}}}} \\ { {1}}&{ {0}} \end{array}} \right|{ { + 1}}\left| {\begin{array}{*{20}{c}} {{ {1 + }}{{ {a}}^{ {2}}}}&{{{ {b}}^{ {2}}}} \\ {{ { - 1}}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\;{ {1 + }}{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + }}{{ {c}}^{ {2}}} \hfill \\ \end{align} \]
= RHS
प्रश्न संख्या 15 तथा 16 मे सही उत्तर चुनिये।
15. यदि \[{ {A}}\] एक \[{ {3 \times 3}}\] कोटी का वर्ग आव्यूह है तो \[{ {|kA|}}\] का मान होगा:
(a) \[{ {k|A|}}\]
(b) \[{{ {k}}^{ {2}}}{ {|\;A|}}\]
(c) \[{{ {k}}^{ {3}}}{ {|\;A|}}\]
(d) \[{ {3k|A|}}\]
उत्तर: \[{ {|A| = }}\left| {\begin{array}{*{20}{l}} { {a}}&{ {b}}&{ {c}} \\ { {d}}&{ {e}}&{ {f}} \\ { {g}}&{ {h}}&{ {i}} \end{array}} \right|\]
\[\begin{align} { {|kA| = }}\left| {\begin{array}{*{20}{l}} {{ {ka}}}&{{ {kb}}}&{{ {kc}}} \\ {{ {kd}}}&{{ {ke}}}&{{ {kf}}} \\ {{ {kg}}}&{{ {kh}}}&{{ {ki}}} \end{array}} \right| \hfill \\ { {|kA| = }}{{ {k}}^{ {3}}}\left| {\begin{array}{*{20}{l}} { {a}}&{ {b}}&{ {c}} \\ { {d}}&{ {e}}&{ {f}} \\ { {g}}&{ {h}}&{ {i}} \end{array}} \right| \hfill \\ { {|kA| = }}{{ {k}}^{ {3}}}{ {|A|}} \hfill \\ \end{align} \]
अतः सही विकल्प (c) है।
16. निम्नलिखित मे से कौन सा कथन सही है।
(a) सारणिक एक वर्ग आव्यूह है।
(b) सारणिक एक आव्यूह से संबंध एक संख्या है।
(c) सारणिक एक वर्ग आव्यूह से संबंध एक संख्या है।
(d) इनमे से कोई नहीं है।
उत्तर: (c) सारणिक एक वर्ग आव्यूह से संबंध संख्या है। क्योंकि सारणिक एक वर्ग आव्यूह का ही निकाला जा सकता है।
प्रश्नावली 4.3
1. निम्नलिखित प्रत्येक मे दिए गए शीर्ष बिन्दुओ वाले त्रिभुज का क्षेत्रफल ज्ञात कीजिए:
(i) \[{ {(1,0), (6,0), (4,3)}}\]
उत्तर: शीर्ष बिन्दुओ \[\left( {{{ {x}}_{ {1}}}{ {,}}{{ {y}}_{ {1}}}} \right){ {, }}\left( {{{ {x}}_{ {2}}}{ {,}}{{ {y}}_{ {2}}}} \right){ {, }}\left( {{{ {x}}_{ {3}}}{ {,}}{{ {y}}_{ {3}}}} \right)\] से होकर जाने वाले त्रिभुज का क्षेत्रफल,
\[{ {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\ {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\ {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}} \end{array}} \right|\]
अभीष्ट त्रिभुज का क्षेत्रफल
\[\begin{align} { {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {1}} \\ { {6}}&{ {0}}&{ {1}} \\ { {4}}&{ {3}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[1(0 - 3) - 0(6 - 4) + 1(18 - 0)]}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 3 + 18] = }}\dfrac{{{ {15}}}}{{ {2}}} \hfill \\ \end{align} \]
(ii) \[{ {(2,7), (1,2), (10,8)}}\]
उत्तर: अभीष्ट त्रिभुज का क्षेत्रफल
\[{ {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}} { {2}}&{ {7}}&{ {1}} \\ { {1}}&{ {1}}&{ {1}} \\ {{ {10}}}&{ {8}}&{ {1}} \end{array}} \right|\]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[2(1 - 8) - 7(1 - 10) + 8(8 - 10)]}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 16 + 63]}} \hfill \\ \end{align} \]
\[{ { = }}\dfrac{{{ {47}}}}{{ {2}}}\]
(iii) \[{ {( - 2, - 3), (3,2), ( - 1, - 8)}}\]
उत्तर: अभीष्ट त्रिभुज का क्षेत्रफल
\[{ {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}} {{ { - 2}}}&{{ { - 3}}}&{ {1}} \\ { {3}}&{ {2}}&{ {1}} \\ {{ { - 1}}}&{{ { - 8}}}&{ {1}} \end{array}} \right|\]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 2(2 + 8) + 3(3 + 1) + ( - 24 + 2)]}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 2(10) + 3(4) + ( - 22)]}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 20 + 12 - 22]}} \hfill \\ { { = - }}\dfrac{{{ {30}}}}{{ {2}}}{ { = - 15}} \hfill \\ \end{align} \]
2. दर्शाइए की बिन्दु \[{ {A(a,b + c), B(b}}{ {.c + a), C(c,a + b)}}\] और संरेख है।
उत्तर: ज्ञात है त्रिभुज के शीर्ष \[{ {A(a,b + c), B(b}}{ {.c + a), C(c,a + b)}}\]
\[{ {\Delta }}\] का क्षेत्रफल = \[{ {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\ {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\ {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}} \end{array}} \right|\]
जहा \[{{ {x}}_{ {1}}}{ { = a, }}{{ {y}}_{ {1}}}{ { = b + c, }}{{ {x}}_{ {2}}}{ { = b, }}{{ {y}}_{ {2}}}{ { = c + a, }}{{ {x}}_{ {3}}}{ { = c, }}{{ {y}}_{ {3}}}{ { = a + b}}\]
\[{ { = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} { {a}}&{{ {b + c}}}&{ {1}} \\ { {b}}&{{ {c + a}}}&{ {1}} \\ { {c}}&{{ {a + b}}}&{ {1}} \end{array}} \right|\,\;\,\;\;\;\;\;\;\,\;\left( {{{ {C}}_{ {1}}} \to {{ {C}}_{ {1}}}{ { + }}{{ {C}}_{ {2}}}} \right)\]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{ {a + b + c}}}&{{ {b + c}}}&{ {1}} \\ {{ {a + b + c}}}&{{ {c + a}}}&{ {1}} \\ {{ {a + b + c}}}&{{ {a + b}}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {(a + b + c)}}\left| {\begin{array}{*{20}{l}} { {1}}&{{ {b + c}}}&{ {1}} \\ { {1}}&{{ {c + a}}}&{ {1}} \\ { {1}}&{{ {a + b}}}&{ {1}} \end{array}} \right| \hfill \\ = \;\dfrac{{ {1}}}{{ {2}}}{ {(a + b + c) \times 0}} \hfill \\ \end{align} \]
\[{ {\Delta }}\] का क्षेत्रफल = \[{ {0}}\]
अतः बिन्दु A, B, C संरेख है।
3. प्रत्येक मे का मान ज्ञात कीजिए यदि त्रिभुज का क्षेत्रफल चार वर्ग इकाई है जहा शरबिदुय निम्नलिखित है:
(i) \[{ {(k,0), (4,0), (0,2)}}\]
उत्तर: त्रिभुज का क्षेत्रफल = \[{ { = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\ {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\ {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}} \end{array}} \right|\]
\[\begin{align} \dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} { {k}}&{ {0}}&{ {1}} \\ { {4}}&{ {0}}&{ {1}} \\ { {0}}&{ {2}}&{ {1}} \end{array}} \right|\; = \; \pm 4 \hfill \\ \left| {\begin{array}{*{20}{l}} { {k}}&{ {0}}&{ {1}} \\ { {4}}&{ {0}}&{ {1}} \\ { {0}}&{ {2}}&{ {1}} \end{array}} \right|{ { = \pm 8}} \hfill \\ { {( - 2)}}\left| {\begin{array}{*{20}{l}} { {k}}&{ {1}} \\ { {4}}&{ {1}} \end{array}} \right|{ { = \pm 8}} \hfill \\ { {k - 4 = \pm 4}} \hfill \\ { {k = 0, 8}} \hfill \\ \end{align} \]
(ii) \[{ {( - 2,0), (0,4), (0,k)}}\]
उत्तर: त्रिभुज का क्षेत्रफल = \[{ { = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\ {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\ {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}} \end{array}} \right|\]
\[\begin{align} = \;\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}} {{ { - 2}}}&{ {0}}&{ {1}} \\ { {0}}&{ {4}}&{ {1}} \\ { {0}}&{ {k}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 2(4 - k) - 0(0 - 0) + 1(0 - 0)}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ { \times ( - 2)(4 - k) = k - 4 = \pm 4}} \hfill \\ { { + ve, k - 4 = 4 ; k = 8}} \hfill \\ { { - ve, k - 4 = 4 ; k = 0}} \hfill \\ \end{align} \]
4. (i) सारणिकों का प्रयोग करके \[{ {(1,2), (3,6)}}\] को मिलाने वाली रेखा का समीकरण ज्ञात कीजिए।
उत्तर: माना कोई बिन्दु \[{ {(x,y)}}\] है
इसलिए त्रिभुज के शीर्ष \[{ {(x,y), (1,2), (3,6)}}\] होंगे।
\[{ {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\ {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\ {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}} \end{array}} \right|\]
जहा \[{{ {x}}_{ {1}}}{ { = x, }}{{ {y}}_{ {1}}}{ { = y, }}{{ {x}}_{ {2}}}{ { = 1, }}{{ {y}}_{ {2}}}{ { = 2, }}{{ {x}}_{ {3}}}{ { = 3, }}{{ {y}}_{ {3}}}{ { = 6}}\]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{c}} { {x}}&{ {y}}&{ {1}} \\ { {1}}&{ {2}}&{ {1}} \\ { {3}}&{ {6}}&{ {1}} \end{array}} \right| \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[x(2 - 6) - y(1 - 3) + 1(6 - 6)]}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[x \times ( - 4) - y( - 2) + 1 \times 0]}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {2}}}{ {[ - 4x + 2y]}} \hfill \\ \end{align} \]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{ {2}}}{ { \times 2( - 2x + y)}} \hfill \\ { { = - 2x + y}} \hfill \\ \end{align} \]
बिन्दु संरेख है।
इसलिए \[{ {\Delta }}\] त्रिभुज का क्षेत्रफल शून्य होगा।
\[\begin{align} { {0 = - 2x + y}} \hfill \\ { {2x - y = 0}} \hfill \\ \end{align} \]
यही अभीष्ट समीकरण है।
(ii) सारणिकों का प्रयोग करके \[{ {(3,1), (9,3)}}\] को मिलाने वाली रेखा का समीकरण ज्ञात कीजिए।
उत्तर: माना बिन्दुओ \[{ {A(3,1), B(9,3)}}\] को मिलाने वाली रेखा पर बिन्दु \[{ {P(x,y)}}\] है। तब बिन्दु A, B, P संरेख है।
\[\begin{align} { {APB = 0}} \hfill \\ \dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} { {3}}&{ {1}}&{\mid { {1}}} \\ { {9}}&{ {3}}&{ {1}} \\ { {x}}&{ {y}}&{ {1}} \end{array}} \right|{ { = 0}} \hfill \\ \left| {\begin{array}{*{20}{l}} { {3}}&{ {1}}&{ {1}} \\ { {9}}&{ {3}}&{ {1}} \\ { {x}}&{ {y}}&{ {1}} \end{array}} \right|{ { = 0}} \hfill \\ \end{align} \]
5. यदि शीर्ष \[{ {(2, - 6), (5,4), (k,4)}}\] वाले त्रिभुज का क्षेत्रफल \[{ {35}}\] वर्ग इकाई हो तो का मान है।
(a) \[{ {12}}\]
(b) \[{ { - 2}}\]
(c) \[{ { - 12, - 2}}\]
(d) \[{ {12, - 2}}\]
उत्तर: दिया है, त्रिभुज के शीर्ष \[{ {(2, - 6), (5,4), (k,4)}}\]
\[{ {\Delta = }}\dfrac{{ {1}}}{{ {2}}}\left| {\begin{array}{*{20}{l}} {{{ {x}}_{ {1}}}}&{{{ {y}}_{ {1}}}}&{ {1}} \\ {{{ {x}}_{ {2}}}}&{{{ {y}}_{ {2}}}}&{ {1}} \\ {{{ {x}}_{ {3}}}}&{{{ {y}}_{ {3}}}}&{ {1}} \end{array}} \right|\]
जहा \[{{ {x}}_{ {1}}}{ { = 2, }}{{ {y}}_{ {1}}}{ { = 6, }}{{ {x}}_{ {2}}}{ { = 5, }}{{ {y}}_{ {2}}}{ { = 4, }}{{ {x}}_{ {3}}}{ { = k, }}{{ {y}}_{ {3}}}{ { = 4}}\]
ज्ञात \[{ {\Delta }}\] का क्षेत्रफल = \[{ { \pm 35}}\]
\[\begin{align} { { \pm 35 = }}\dfrac{{ {1}}}{{ {2}}}{ {[2(4 - 4) + 6(5 - k) + k(120 - 4k)]}} \hfill \\ { { \pm 35 = }}\dfrac{{ {1}}}{{ {2}}}{ {[2 \times 0 + 6(5 - k) + 1(20 - 4k)]}} \hfill \\ { { \pm 70 = 6(5 - k) + 20 - 4k}} \hfill \\ { { \pm 70 = 30 - 6k + 20 - 4k}} \hfill \\ { { \pm 70 = 50 - 10k}} \hfill \\ { { \pm 70 = 5 - k}} \hfill \\ { { + ve, 7 = 5 - k ; k = 5 - 7 = - 2}} \hfill \\ { { - ve, - 7 = 5 - k ; - 12 = - k ; k = 12}} \hfill \\ \end{align} \]
अतः \[{ {k = 12, - 2}}\]
अतः विकल्प (d) सही है।
प्रश्नावली 4.4
निम्नलिखित सारणिकों के अवयवों के उपसारणिक एवं सहखंड लिखिए:
1. (i) \[\left| {\begin{array}{*{20}{c}} { {2}}&{{ { - 4}}} \\ { {0}}&{ {3}} \end{array}} \right|\]
उत्तर: उपसारणिक \[{{ {M}}_{11}}{ { = 3, }}{{ {M}}_{12}}{ { = 0, }}{{ {M}}_{21}}{ { = - 4, }}{{ {M}}_{22}}{ { = 2}}\]
तथा सहखंड \[{{ {A}}_{11}}{ { = 3, }}{{ {A}}_{12}}{ { = 0, }}{{ {A}}_{21}}{ { = - ( - 4), }}{{ {A}}_{22}}{ { = 2}}\]
(ii) \[\left| {\begin{array}{*{20}{l}} { {a}}&{ {c}} \\ { {b}}&{ {d}} \end{array}} \right|\]
उत्तर: उपसारणिक \[{{ {M}}_{11}}{ { = d, }}{{ {M}}_{12}}{ { = b, }}{{ {M}}_{21}}{ { = c, }}{{ {M}}_{22}}{ { = a}}\]
तथा सहखंड \[{{ {A}}_{11}}{ { = d, }}{{ {A}}_{12}}{ { = - b, }}{{ {A}}_{21}}{ { = - c, }}{{ {A}}_{22}}{ { = a}}\]
2. (i) \[\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right|\]
उत्तर: उपसारणिक
\[\begin{align} {{ {M}}_{{ {11}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 1 - 0 = 1 , }}{{ {M}}_{{ {12}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 0}} \hfill \\ {{ {M}}_{{ {13}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {0}}&{ {1}} \\ { {0}}&{ {0}} \end{array}} \right|{ { = 0 , }}{{ {M}}_{{ {21}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 0}} \hfill \\ {{ {M}}_{{ {22}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 1 , }}{{ {M}}_{{ {23}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right|{ { = 0}} \hfill \\ {{ {M}}_{{ {31}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 0 , }}{{ {M}}_{{ {32}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right|{ { = 0}} \hfill \\ {{ {M}}_{{ {31}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 0}} \hfill \\ \end{align} \]
तथा सहखंड
\[\begin{align} {{ {A}}_{{ {11}}}}{ { = ( - 1}}{{ {)}}^{{ {1 + 1}}}}{{ {M}}_{{ {11}}}}{ { = 1 , }}{{ {A}}_{{ {12}}}}{ { = ( - 1}}{{ {)}}^{{ {1 + 2}}}}{{ {M}}_{{ {12}}}}{ { = ( - 1) \times 0 = 0}} \hfill \\ {{ {A}}_{{ {13}}}}{ { = ( - 1}}{{ {)}}^{{ {1 + 3}}}}{{ {M}}_{{ {12}}}}{ { = 1 \times 0 = 0 , }}{{ {A}}_{{ {21}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 1}}}}{{ {M}}_{{ {21}}}}{ { = 0}} \hfill \\ \end{align} \]
\[\begin{align} {{ {A}}_{{ {22}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 2}}}}{{ {M}}_{{ {22}}}}{ { = 1 , }}{{ {A}}_{{ {23}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 3}}}}{{ {M}}_{{ {23}}}}{ { = 0}} \hfill \\ {{ {A}}_{{ {31}}}}{ { = ( - 1}}{{ {)}}^{{ {3 + 1}}}}{{ {M}}_{{ {31}}}}{ { = 0 , }}{{ {A}}_{{ {32}}}}{ { = ( - 1}}{{ {)}}^{{ {3 + 2}}}}{{ {M}}_{{ {32}}}}{ { = 0}} \hfill \\ {{ {A}}_{{ {33}}}}{ { = ( - 1}}{{ {)}}^{{ {3 + 3}}}}{{ {M}}_{{ {33}}}}{ { = 1}}{ {.1 = 1}} \hfill \\ \end{align} \]
(ii) \[\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {4}} \\ { {3}}&{ {5}}&{{ { - 1}}} \\ { {0}}&{ {1}}&{ {2}} \end{array}} \right|\]
उत्तर: उपसारणिक
\[\begin{align} {{ {M}}_{{ {11}}}}{ { = }}\left| {\begin{array}{*{20}{c}} { {5}}&{{ { - 1}}} \\ { {1}}&{ {2}} \end{array}} \right|{ { = 5 \times 2 - 1(1) = 11 , }}{{ {M}}_{{ {12}}}}{ { = }}\left| {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}} \\ { {0}}&{ {2}} \end{array}} \right|{ { = 3 \times 2 - 0 = 6}} \hfill \\ {{ {M}}_{{ {13}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {3}}&{ {5}} \\ { {0}}&{ {1}} \end{array}} \right|{ { = 3 \times 1 - 0 = 3 , }}{{ {M}}_{{ {21}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {0}}&{ {4}} \\ { {1}}&{ {2}} \end{array}} \right|{ { = 0 \times 2 - 1 \times 4 = - 4}} \hfill \\ {{ {M}}_{{ {22}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {4}} \\ { {0}}&{ {2}} \end{array}} \right|{ { = 1 \times 2 - 0 = 2 , }}{{ {M}}_{{ {23}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {2}} \end{array}} \right|{ { = 1 \times 1 = 1}} \hfill \\ {{ {M}}_{{ {32}}}}{ { = }}\left| {\begin{array}{*{20}{c}} { {1}}&{ {4}} \\ { {3}}&{{ { - 1}}} \end{array}} \right|{ { = 1 \times - (1) - 3 \times 4 = 13 , }}{{ {M}}_{{ {33}}}}{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {3}}&{ {5}} \end{array}} \right|{ { = 5 - 0 = 5}} \hfill \\ \end{align} \]
तथा सहखंड
\[\begin{align} {{ {A}}_{{ {11}}}}{ { = ( - 1}}{{ {)}}^{{ {1 + 1}}}}{{ {M}}_{{ {11}}}}{ { = 11 , }}{{ {A}}_{{ {12}}}}{ { = ( - 1}}{{ {)}}^{{ {1 + 2}}}}{{ {M}}_{{ {12}}}}{ { = ( - 1) \times 6 = - 6}} \hfill \\ {{ {A}}_{{ {13}}}}{ { = ( - 1}}{{ {)}}^{{ {1 + 3}}}}{{ {M}}_{{ {12}}}}{ { = 1}}{ {.3 = 3 , }}{{ {A}}_{{ {21}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 1}}}}{{ {M}}_{{ {21}}}}{ { = ( - 1) \times ( - 4) = 4}} \hfill \\ \end{align} \] \[\begin{align} {{ {A}}_{{ {22}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 2}}}}{{ {M}}_{{ {22}}}}{ { = 1 \times 2 = 2 , }}{{ {A}}_{{ {23}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 3}}}}{{ {M}}_{{ {23}}}}{ { = ( - 1)1 = - 1}} \hfill \\ {{ {A}}_{{ {31}}}}{ { = ( - 1}}{{ {)}}^{{ {3 + 1}}}}{{ {M}}_{{ {31}}}}{ { = 1}}{ {.( - 20) = - 20 , }}{{ {A}}_{{ {32}}}}{ { = ( - 1}}{{ {)}}^{{ {3 + 2}}}}{{ {M}}_{{ {32}}}}{ { = ( - 1)( - 13) = 13}} \hfill \\ {{ {A}}_{{ {33}}}}{ { = ( - 1}}{{ {)}}^{{ {3 + 3}}}}{{ {M}}_{{ {33}}}}{ { = 1}}{ {.5 = 1}} \hfill \\ \end{align} \]
3. दूसरी पंक्ति के अवयवों के सहखंडों का प्रयोग करके का मान ज्ञात कीजिए।
उत्तर: \[{{ {A}}_{{ {21}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 1}}}}\left| {\begin{array}{*{20}{l}} { {3}}&{ {8}} \\ { {2}}&{ {3}} \end{array}} \right|{ { = ( - 1) \times [3 \times 3 - 2 \times 8] = 7}}\]
\[\begin{align} {{ {A}}_{{ {22}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 2}}}}\left| {\begin{array}{*{20}{l}} { {5}}&{ {8}} \\ { {1}}&{ {3}} \end{array}} \right|{ { = 1[5 \times 3 - 2 \times 8] = 7}} \hfill \\ {{ {A}}_{{ {23}}}}{ { = ( - 1}}{{ {)}}^{{ {2 + 3}}}}\left| {\begin{array}{*{20}{l}} { {5}}&{ {3}} \\ { {1}}&{ {2}} \end{array}} \right|{ { = ( - 1)[5 \times 2 - 3 \times 1] = - 7}} \hfill \\ { {\Delta = }}{{ {a}}_{{ {21}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {22}}}}{{ {A}}_{{ {22}}}}{ { + }}{{ {a}}_{{ {23}}}}{{ {A}}_{{ {23}}}} \hfill \\ { { = 2 \times 7 + 0 \times 7 + 1 \times ( - 7)}} \hfill \\ { { = 14 - 7 = 7}} \hfill \\ \end{align} \]
4. तीसरे स्तम्भ के अवयवों के सहखंडों का प्रयोग करके \[{ {\Delta = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {x}}&{{ {yz}}} \\ { {1}}&{ {y}}&{{ {zx}}} \\ { {1}}&{ {2}}&{{ {xy}}} \end{array}} \right|\] का मान ज्ञात कीजिए।
उत्तर: \[{{ {A}}_{{ {13}}}}{ { = - }}{{ {1}}^{{ {1 + 3}}}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {y}} \\ { {1}}&{ {z}} \end{array}} \right|{ { = (1)(z - y) = (z - y)}}\]
\[{{ {A}}_{{ {23}}}}{ { = - }}{{ {1}}^{{ {2 + 3}}}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {x}} \\ { {1}}&{ {z}} \end{array}} \right|{ { = (1)(z - x) = - (x - z)}}\]
\[\begin{align} {{ {A}}_{{ {33}}}}{ { = - }}{{ {1}}^{{ {1 + 3}}}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {x}} \\ { {1}}&{ {y}} \end{array}} \right|{ { = (1)(y - x) = (y - x)}} \hfill \\ { {\Delta = }}{{ {a}}_{{ {13}}}}{{ {A}}_{{ {13}}}}{ { + }}{{ {a}}_{{ {23}}}}{{ {A}}_{{ {23}}}}{ { + }}{{ {a}}_{{ {33}}}}{{ {A}}_{{ {33}}}} \hfill \\ { { = yz(z - y) + zx(x - z) + xy(y - x)}} \hfill \\ { { = y}}{{ {z}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}{ {z + z}}{{ {x}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}{ {x + x}}{{ {y}}^{ {2}}}{ { - }}{{ {x}}^{ {2}}}{ {y}} \hfill \\ { { = z}}{{ {x}}^{ {2}}}{ { - }}{{ {x}}^{ {2}}}{ {y + x}}{{ {y}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}{ {y + y}}{{ {z}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}{ {z}} \hfill \\ { { = }}{{ {x}}^{ {2}}}{ {((z - y) + z(y - z)(y + z) + yz(z - y)}} \hfill \\ { { = (z - y)}}\left[ {{{ {x}}^{ {2}}}{ { - x(y + z) + yz}}} \right] \hfill \\ { { = (z - y)}}\left[ {{{ {x}}^{ {2}}}{ { - xy - xz + yz}}} \right] \hfill \\ { { = (z - y)[x(x - y) - z(x - y)]}} \hfill \\ { { = (z - y)(x - y)(x - z)}} \hfill \\ { { = (x - y)(y - z)(z - x)}} \hfill \\ \end{align} \]
5. यदि \[{ {\Delta = }}\left| {\begin{array}{*{20}{l}} {{{ {a}}_{{ {11}}}}}&{{{ {a}}_{{ {12}}}}}&{{{ {a}}_{{ {13}}}}} \\ {{{ {a}}_{{ {21}}}}}&{{{ {a}}_{{ {22}}}}}&{{{ {a}}_{{ {23}}}}} \\ {{{ {a}}_{{ {31}}}}}&{{{ {a}}_{{ {32}}}}}&{{{ {a}}_{{ {33}}}}} \end{array}} \right|\] और \[{{ {a}}_{ {y}}}\] का सहखंड \[{{ {A}}_{ {y}}}\] हो तो \[{ {\Delta }}\] शन का मान निम्नलिखित रूप मे व्यक्त किया जाता है:
(a) \[{{ {a}}_{{ {11}}}}{ { \times }}{{ {A}}_{{ {31}}}}{ { + }}{{ {a}}_{{ {12}}}}{{ {A}}_{{ {32}}}}{ { + }}{{ {a}}_{{ {13}}}}{{ {A}}_{{ {13}}}}\]
(b) \[{{ {a}}_{{ {11}}}}{ { \times }}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {12}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {13}}}}{{ {A}}_{{ {31}}}}\]
(c) \[{{ {a}}_{{ {21}}}}{ { \times }}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {22}}}}{{ {A}}_{{ {12}}}}{ { + }}{{ {a}}_{{ {23}}}}{{ {A}}_{{ {13}}}}\]
(d) \[{{ {a}}_{{ {11}}}}{ { \times }}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {21}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {31}}}}{{ {A}}_{{ {31}}}}\]
उत्तर: \[{ {\Delta }}\] = किसी पंक्ति अथवा स्तम्भ के अवयवों तथा उनके संगत महखंडों के गुणा का योग
\[{{ {C}}_{ {1}}}\] = स्तम्भ के अवयवों
= \[{{ {A}}_{{ {11}}}}{ {,}}\;{{ {A}}_{{ {21}}}}{ {,}}\;{{ {A}}_{{ {31}}}}\]
\[ \Rightarrow \;{{ {a}}_{{ {11}}}}{{ {A}}_{{ {11}}}}{ { + }}{{ {a}}_{{ {21}}}}{{ {A}}_{{ {21}}}}{ { + }}{{ {a}}_{{ {31}}}}{{ {A}}_{{ {31}}}}\]
अतः विकल्प (d) सही है।
प्रश्नावली 4.5
प्रश्न 1 और 2 मे प्रत्येक आव्यूह का सहखंडज ज्ञात कीजिए।
1. \[\left[ \begin{align} 1\,\,\,\,\,2 \hfill \\ 3\;\,\,4 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {2}} \\ { {3}}&{ {4}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{c}} { {4}}&{{ { - 2}}} \\ {{ { - 3}}}&{ {1}} \end{array}} \right]\]
2. \[\left[ \begin{align} 1\,\;\,\;\,\;\; - 1\;\;\;\;\;\;\;2 \hfill \\ 2\;\;\;\;\;\;\;3\;\;\;\;\;\;\;\;5 \hfill \\ - 2\;\;\;\,\;\;\;0\;\;\;\;\;\;1 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {2}}&{ {3}}&{ {5}} \\ {{ { - 2}}}&{ {0}}&{ {1}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}}&{{ { - 11}}} \\ {{ { - 12}}}&{ {5}}&{{ { - 1}}} \\ { {6}}&{ {2}}&{ {5}} \end{array}} \right]\]
प्रश्न 3 और 4 मे सत्यापित कीजिए की \[{ {A(adjA) = (adjA) \times A = |A|}}{ {.I}}\] है।
3. \[\left[ \begin{align} 2\;\;\;\;\;\;\;3 \hfill \\ - 4\;\;\; - 6 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {3}} \\ {{ { - 4}}}&{{ { - 6}}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{c}} {{ { - 6}}}&{{ { - 3}}} \\ { {4}}&{ {2}} \end{array}} \right]\]
\[\begin{align} { {A(adjA) = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {3}} \\ {{ { - 4}}}&{{ { - 6}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{ { - 6}}}&{{ { - 3}}} \\ { {4}}&{ {2}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 12 + 12}}}&{{ { - 6 + 6}}} \\ {{ {24 - 24}}}&{{ {12 - 12}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} { {(adjA)A = }}\left[ {\begin{array}{*{20}{c}} {{ { - 6}}}&{{ { - 3}}} \\ { {4}}&{ {2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {2}}&{ {3}} \\ {{ { - 4}}}&{{ { - 6}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 12 + 12}}}&{{ { - 18 + 18}}} \\ {{ {8 - 8}}}&{{ {12 - 12}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} { {|A| I = 0 \times }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ { {A(adjA) = (adjA)A = |A| I = }}\left[ {\begin{array}{*{20}{c}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ \end{align} \]
4. \[\left[ \begin{align} 1\,\;\,\;\,\;\; - 1\;\;\;\;\;\;\;2 \hfill \\ 3\;\;\;\;\;\;\;\;0\;\;\;\;\; - 2 \hfill \\ 1\;\;\;\,\;\;\;\;0\;\;\;\;\;\;\;\;3 \hfill \\ \end{align} \right]\]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {3}}&{ {0}}&{{ { - 2}}} \\ { {1}}&{ {0}}&{ {3}} \end{array}} \right]\]
\[\begin{align} { {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {0}}&{ {3}}&{ {2}} \\ {{ { - 11}}}&{ {1}}&{ {8}} \\ { {0}}&{{ { - 1}}}&{ {3}} \end{array}} \right] \hfill \\ { {A(adjA) = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {3}}&{ {0}}&{{ { - 2}}} \\ { {1}}&{ {0}}&{ {3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {0}}&{ {3}}&{ {2}} \\ {{ { - 11}}}&{ {1}}&{ {8}} \\ { {0}}&{{ { - 1}}}&{ {3}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ {11}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {11}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {11}}} \end{array}} \right] \hfill \\ { {(adjA)A = }}\left[ {\begin{array}{*{20}{c}} { {0}}&{ {3}}&{ {2}} \\ {{ { - 11}}}&{ {1}}&{ {8}} \\ { {0}}&{{ { - 1}}}&{ {3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {3}}&{ {0}}&{{ { - 2}}} \\ { {1}}&{ {0}}&{ {3}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {11}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {11}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {11}}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} { {|A| I = 11 }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {11}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {11}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {11}}} \end{array}} \right] \hfill \\ { {A(adjA) = (adjA)A = |A| I = }}\left[ {\begin{array}{*{20}{c}} {{ {11}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {11}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {11}}} \end{array}} \right] \hfill \\ \end{align} \]
प्रश्न 5 से 11 मे दिए गए प्रत्येक आवयहो के व्युत्क्रम ज्ञात कीजिए।
5. \[\left[ \begin{align} 2\;\;\;\,\;\;\, - 2 \hfill \\ 4\;\,\;\;\;\;\;\;\,3 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 2}}} \\ { {4}}&{ {3}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {2}} \\ {{ { - 4}}}&{ {2}} \end{array}} \right]\]
\[{ {|A| = 6 + 8 = 14}}\; \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {14}}}}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {2}} \\ {{ { - 4}}}&{ {2}} \end{array}} \right]\]
6. \[\left[ \begin{align} - 1\;\;\;\,\;\;\,\;\;5 \hfill \\ - 3\;\,\;\;\;\;\;\;\,2 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{l}} {{ { - 1}}}&{ {5}} \\ {{ { - 3}}}&{ {2}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} { {2}}&{{ { - 5}}} \\ { {3}}&{{ { - 1}}} \end{array}} \right]\]
\[{ {|A| = - 2 + 15 = 13 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{l}} { {2}}&{{ { - 5}}} \\ { {3}}&{{ { - 1}}} \end{array}} \right]\]
7. \[\left[ \begin{align} 1\;\;\;\,\;\;\,\;\;2\;\;\,\;\;\,\;\;3 \hfill \\ 0\;\;\,\;\;\;\;\;2\;\;\;\;\;\;\,\;4 \hfill \\ 0\;\,\;\;\;\;\;\;\,0\;\;\;\;\;\;\;5 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {2}}&{ {3}} \\ { {0}}&{ {2}}&{ {4}} \\ { {0}}&{ {0}}&{ {5}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {10}}}&{{ { - 10}}}&{ {2}} \\ { {0}}&{ {5}}&{{ { - 4}}} \\ { {0}}&{ {0}}&{ {2}} = \end{array}} \right]\]
\[{ {|A| = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) = 10 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}} {{ {10}}}&{{ { - 10}}}&{ {2}} \\ { {0}}&{ {5}}&{{ { - 4}}} \\ { {0}}&{ {0}}&{ {2}} \end{array}} \right] \hfill \\ \end{align} \]
8. \[\left[ \begin{align} 1\;\;\;\,\;\;\,\;\;0\;\;\,\;\;\,\;\;0 \hfill \\ 3\;\;\,\;\;\;\;\;3\;\;\;\;\;\;\,\;0 \hfill \\ 5\;\,\;\;\;\;\;\;\,2\;\;\;\;\;\; - 1 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {3}}&{ {3}}&{ {0}} \\ { {5}}&{ {2}}&{{ { - 1}}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 3}}}&{ {0}}&{ {0}} \\ { {3}}&{{ { - 1}}}&{ {0}} \\ {{ { - 9}}}&{{ { - 2}}}&{ {3}} \end{array}} \right]\]
\[{ {|A| = 1( - 3 - 0) - 0( - 3 - 0) + 0(6 - 15) = - 3 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ { - 3}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 3}}}&{ {0}}&{ {0}} \\ { {3}}&{{ { - 1}}}&{ {0}} \\ {{ { - 9}}}&{{ { - 2}}}&{ {3}} \end{array}} \right] \hfill \\ \end{align} \]
9. \[\left[ \begin{align} 2\;\;\;\,\;\;\,\;\;\;\,\;\;1\;\,\;\;\,\;\;\;\;\;3 \hfill \\ 4\;\;\,\;\;\;\;\;\;\; - 1\;\;\;\;\,\;\;\;0 \hfill \\ - 7\;\,\;\;\;\;\;\;\;\;\,2\;\;\;\;\;\,\;\;1 \hfill \\ \end{align} \right]\]
उत्तर: यह \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {1}}&{ {3}} \\ { {4}}&{{ { - 1}}}&{ {0}} \\ {{ { - 7}}}&{ {2}}&{ {1}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {5}}&{ {3}} \\ {{ { - 4}}}&{{ {23}}}&{{ {12}}} \\ { {1}}&{{ { - 11}}}&{{ { - 6}}} \end{array}} \right]\]
\[{ {|A| = 2( - 1 - 0) - 1(4 - 0) + 3(8 - 7) = - 3 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ { - 3}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {5}}&{ {3}} \\ {{ { - 4}}}&{{ {23}}}&{{ {12}}} \\ { {1}}&{{ { - 11}}}&{{ { - 6}}} \end{array}} \right] \hfill \\ \end{align} \]
10. \[\left[ \begin{align} 1\;\;\;\,\;\;\,\;\; - 1\;\,\;\;\,\;\;\;\;\;2 \hfill \\ 0\;\;\,\;\;\;\;\;\;\;2\;\;\;\;\,\;\;\; - 3 \hfill \\ 3\;\,\;\;\;\;\;\, - 2\;\;\;\;\;\,\;\;4 \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {0}}&{ {2}}&{{ { - 3}}} \\ { {3}}&{{ { - 2}}}&{ {4}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {0}}&{{ { - 1}}} \\ {{ { - 9}}}&{{ { - 2}}}&{ {3}} \\ {{ { - 6}}}&{{ { - 1}}}&{ {2}} \end{array}} \right]\]
\[{ {|A| = 1(8 - 6) + 1(0 + 9) + 2(0 - 6) = - 1 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ { - 1}}}}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {0}}&{{ { - 1}}} \\ {{ { - 9}}}&{{ { - 2}}}&{ {3}} \\ {{ { - 6}}}&{{ { - 1}}}&{ {2}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 2}}}&{ {0}}&{ {1}} \\ { {9}}&{ {2}}&{{ { - 3}}} \\ { {6}}&{ {1}}&{{ { - 2}}} \end{array}} \right] \hfill \\ \end{align} \]
11. \[\left[ \begin{align} { {1}}\;\;\;\,\;\;\,\;\;{ {0}}\;\,\;\;\,\;\;\;\;\;\;\;\;\;\,{ {0}} \hfill \\ { {0}}\;\;\,\;\;\;\;{ {cos\alpha }}\;\;\;\;\,\;\;\;{ {sin\alpha }} \hfill \\ { {0}}\;\,\;\;\;\;\,{ {sin\alpha }}\;\;\;\;\;\,\;\;{ {cos\alpha }} \hfill \\ \end{align} \right]\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{{ {cos\alpha }}}&{{ {sin\alpha }}} \\ { {0}}&{{ {sin\alpha }}}&{{ { - cos\alpha }}} \end{array}} \right]\]
इसलिए \[{ {adjA = }}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{{ { - cos\alpha }}}&{{ { - sin\alpha }}} \\ { {0}}&{{ { - sin\alpha }}}&{{ {cos\alpha }}} \end{array}} \right]\]
\[{ {|A| = 1}}\left( {{ { - co}}{{ {s}}^{ {2}}}{ {\alpha - si}}{{ {n}}^{ {2}}}{ {\alpha }}} \right){ { + 0(0 - 0) + 0(0 - 0) = - 1 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {31}}}}} \\ {{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {32}}}}} \\ {{{ {A}}_{{ {13}}}}}&{{{ {A}}_{{ {23}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ { - 1}}}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{{ { - cos\alpha }}}&{{ { - sin\alpha }}} \\ { {0}}&{{ { - sin\alpha }}}&{{ {cos\alpha }}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {cos\alpha }}}&{{ {sin\alpha }}} \\ { {0}}&{{ {sin\alpha }}}&{{ { - cos\alpha }}} \end{array}} \right] \hfill \\ \end{align} \]
12. यदि \[{ {A}}\;{ { = }}\;\left[ \begin{align} { {3}}\;\;\;\,\;\;\,\;\;{ {7}} \hfill \\ { {2}}\;\,\;\;\;\;\;\;\,{ {5}} \hfill \\ \end{align} \right]\] और \[{ {B}}\;{ { = }}\;\left[ \begin{align} { {6}}\;\;\;\,\;\;\,\;\;{ {8}} \hfill \\ { {7}}\;\,\;\;\;\;\;\;\,{ {9}} \hfill \\ \end{align} \right]\] है तो सत्यापित कीजिए की \[{{ {(AB)}}^{{ { - 1}}}}\,{ { = }}\,{{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}}\]
उत्तर: यहा \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {3}}&{ {7}} \\ { {2}}&{ {5}} \end{array}} \right]\]
\[{ {|A| = 15 - 14 = 1 }} \ne { { 0 }} \to { { }}{{ {A}}^{{ { - 1}}}}\] का अस्तित्व है।
\[{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {(adjA) = }}\dfrac{{ {1}}}{{ {1}}}\left[ {\begin{array}{*{20}{c}} { {5}}&{{ { - 7}}} \\ {{ { - 2}}}&{ {3}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {5}}&{{ { - 7}}} \\ {{ { - 2}}}&{ {3}} \end{array}} \right]\]
यहा \[{ {B = }}\left[ {\begin{array}{*{20}{l}} { {6}}&{ {8}} \\ { {7}}&{ {9}} \end{array}} \right]\]
\[{ {|B| = 54 - 56 = - 2 }} \ne { { 0 }} \to { { }}{{ {B}}^{{ { - 1}}}}\] का अस्तित्व है।
\[{{ {B}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|B|}}}}{ {(adjB) = }}\dfrac{{ {1}}}{{{ { - 2}}}}\left[ {\begin{array}{*{20}{c}} { {9}}&{{ { - 8}}} \\ {{ { - 7}}}&{ {6}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{ {9}}}{{ {2}}}}&{ {4}} \\ {\dfrac{{ {7}}}{{ {2}}}}&{{ { - 3}}} \end{array}} \right]\]
\[\begin{align} {{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}}{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{ {9}}}{{ {2}}}}&{ {4}} \\ {\dfrac{{ {7}}}{{ {2}}}}&{{ { - 3}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {5}}&{{ { - 7}}} \\ {{ { - 2}}}&{ {3}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{{ {45}}}}{{ {2}}}{ { - 8}}}&{\dfrac{{{ {63}}}}{{ {2}}}{ { + 12}}} \\ {\dfrac{{{ {35}}}}{{ {2}}}{ { + 6}}}&{{ { - }}\dfrac{{{ {49}}}}{{ {2}}}{ { - 9}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{{ {81}}}}{{ {2}}}}&{\dfrac{{{ {87}}}}{{ {2}}}} \\ {\dfrac{{{ {47}}}}{{ {2}}}}&{{ { - }}\dfrac{{{ {67}}}}{{ {2}}}} \end{array}} \right] \hfill \\ { {AB = }}\left[ {\begin{array}{*{20}{l}} { {3}}&{ {7}} \\ { {2}}&{ {5}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {6}}&{ {8}} \\ { {7}}&{ {9}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} {{ {18 + 49}}}&{{ {24 + 63}}} \\ {{ {12 + 35}}}&{{ {16 + 45}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} {{ {67}}}&{{ {87}}} \\ {{ {47}}}&{{ {61}}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ {|AB| = 4087 - 4089 = - 2 }} \ne { { 0 }} \to { { (AB}}{{ {)}}^{{ { - 1}}}}\] का अस्तित्व है।
\[\begin{align} {{ {(AB)}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|AB|}}}}{ {(adjAB) = }}\dfrac{{ {1}}}{{{ { - 2}}}}\left[ {\begin{array}{*{20}{c}} {{ {61}}}&{{ { - 87}}} \\ {{ { - 47}}}&{{ {67}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{{ {61}}}}{{ {2}}}}&{\dfrac{{{ {87}}}}{{ {2}}}} \\ {\dfrac{{{ {47}}}}{{ {2}}}}&{{ { - }}\dfrac{{{ {67}}}}{{ {2}}}} \end{array}} \right] \hfill \\ {{ {(AB)}}^{{ { - 1}}}}{ { = }}{{ {B}}^{{ { - 1}}}}{ {\;}}{{ {A}}^{{ { - 1}}}} \hfill \\ \end{align} \]
13. यदि \[{ {A}}\;{ { = }}\;\left[ \begin{align} 3\;\;\;\,\;\;\,\;\;\;1 \hfill \\ - 1\;\,\;\;\;\;\;\;2 \hfill \\ \end{align} \right]\] है तो दर्शाइए की \[{{ {A}}^{ {2}}}{ { - 5A + 7I}}\;{ { = }}\;{ {0}}\] है इसकी सहायता से \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए।
उत्तर: LHS = \[{{ {A}}^{ {2}}}{ { - 5A + 7I = AA - 5A + 7I}}\]
\[\begin{align} { { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}} \\ {{ { - 1}}}&{ {2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}} \\ {{ { - 1}}}&{ {2}} \end{array}} \right]{ { - 5}}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}} \\ {{ { - 1}}}&{ {2}} \end{array}} \right]{ { + 7}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ {9 - 1}}}&{{ {3 + 2}}} \\ {{ { - 3 - 2}}}&{{ { - 1 + 4}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} {{ {15}}}&{ {5}} \\ {{ { - 5}}}&{{ {10}}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} { {7}}&{ {0}} \\ { {0}}&{ {7}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ {8 - 15 + 7}}}&{{ {5 - 5 + 0}}} \\ {{ { - 5 + 5 + 0}}}&{{ {3 - 10 + 7}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ { { = 0}} \hfill \\ \end{align} \]
= RHS
\[\begin{align} {{ {A}}^{ {2}}}{ { - 5A + 7I = 0}} \hfill \\ {{ {A}}^{ {2}}}{ { - 5A = - 7I}} \hfill \\ { {AA}}{{ {A}}^{{ { - 1}}}}{ { - 5A}}{{ {A}}^{{ { - 1}}}}{ { = - 7I}}{{ {A}}^{{ { - 1}}}} \hfill \\ { {7}}{{ {A}}^{{ { - 1}}}}{ { = 5I - AI = 5}}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}} \\ {{ { - 1}}}&{ {2}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{l}} { {5}}&{ {0}} \\ { {0}}&{ {5}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}} \\ {{ { - 1}}}&{ {2}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}} \\ { {1}}&{ {3}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{ {7}}}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}} \\ { {1}}&{ {3}} \end{array}} \right] \hfill \\ \end{align} \]
14. आव्यूह \[{ {A}}\;{ { = }}\;\left[ \begin{align} 3\;\;\;\,\;\;\,\;\;\;2 \hfill \\ 1\;\,\;\;\;\;\;\;\;\;1 \hfill \\ \end{align} \right]\] के लिए और ऐसी संख्या ज्ञात कीजिए ताकि \[{{ {A}}^{ {2}}}{ { + aA + bI}}\;{ { = }}\;{ {0}}\] हो।
उत्तर: \[{{ {A}}^{ {2}}}{ { + aA + bI}}\;{ { = }}\;{ {0}}\]
\[\begin{align} { { = }}\left[ {\begin{array}{*{20}{l}} { {3}}&{ {2}} \\ { {1}}&{ {1}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {3}}&{ {2}} \\ { {1}}&{ {1}} \end{array}} \right]{ { + a}}\left[ {\begin{array}{*{20}{l}} { {3}}&{ {2}} \\ { {1}}&{ {1}} \end{array}} \right]{ { + b}}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}} \\ { {0}}&{ {1}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{l}} {{ {9 + 2}}}&{{ {6 + 2}}} \\ {{ {3 + 1}}}&{{ {2 + 1}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} {{ {3a}}}&{{ {2a}}} \\ { {a}}&{ {a}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} { {b}}&{ {0}} \\ { {0}}&{ {b}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{l}} {{ {11 + 3a + b}}}&{{ {8 + 2a + 0}}} \\ {{ {4 + a + 0}}}&{{ {3 + a + b}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}} \\ { {0}}&{ {0}} \end{array}} \right] \hfill \\ { {4 + a = 0}} \hfill \\ { {a = - 4}} \hfill \\ { {3 + a + b = 0}} \hfill \\ { {3 - 4 + b = 0}} \hfill \\ { {b = 1}} \hfill \\ { {a = - 4, b = 1}} \hfill \\ \end{align} \]
15. आव्यूह \[{ {A}}\;{ { = }}\;\left[ \begin{align} 1\;\;\;\,\;\;\;1\;\,\;\;\;\;\;1 \hfill \\ 1\;\;\;\;\;\;\,\;2\;\;\; - 3 \hfill \\ 2\;\,\;\;\; - 1\;\;\;\;\;\;3 \hfill \\ \end{align} \right]\] के लिए दर्शाइए की \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A + 11 I}}\;{ { = }}\;{ {0}}\] है। इसकी सहायता से \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए।
उत्तर: \[{ {A}}\;{ { = }}\;\left[ \begin{align} 1\;\;\;\,\;\;\;1\;\,\;\;\;\;\;1 \hfill \\ 1\;\;\;\;\;\;\,\;2\;\;\; - 3 \hfill \\ 2\;\,\;\;\; - 1\;\;\;\;\;\;3 \hfill \\ \end{align} \right]\]
\[\begin{align} {{ {A}}^2}{ { = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {1}}&{ {2}}&{{ { - 3}}} \\ { {2}}&{{ { - 1}}}&{ {3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {1}}&{ {2}}&{{ { - 3}}} \\ { {2}}&{{ { - 1}}}&{ {3}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {2}}&{ {1}} \\ {{ { - 3}}}&{ {8}}&{{ { - 14}}} \\ { {7}}&{{ { - 3}}}&{{ {14}}} \end{array}} \right] \hfill \\ {{ {A}}^3}{ { = }}{{ {A}}^2}{ {A = }}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {2}}&{ {1}} \\ {{ { - 3}}}&{ {8}}&{{ { - 14}}} \\ { {7}}&{{ { - 3}}}&{{ {14}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {1}}&{ {2}}&{{ { - 3}}} \\ { {2}}&{{ { - 1}}}&{ {3}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} { {8}}&{ {7}}&{ {1}} \\ {{ { - 23}}}&{{ {27}}}&{{ { - 69}}} \\ {{ {32}}}&{{ { - 13}}}&{{ {58}}} \end{array}} \right] \hfill \\ \end{align} \]
LHS = \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A + 11 I}}\;{ { = }}\;{ {0}}\]
\[\begin{align} { { = }}\left[ {\begin{array}{*{20}{c}} { {8}}&{ {7}}&{ {1}} \\ {{ { - 23}}}&{{ {27}}}&{{ { - 69}}} \\ {{ {32}}}&{{ { - 13}}}&{{ {58}}} \end{array}} \right]{ { - 6}}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {2}}&{ {1}} \\ {{ { - 3}}}&{ {8}}&{{ { - 14}}} \\ { {7}}&{{ { - 3}}}&{{ {14}}} \end{array}} \right]{ { + 5}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {1}}&{ {2}}&{{ { - 3}}} \\ { {2}}&{{ { - 1}}}&{ {3}} \end{array}} \right]{ { + 11}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} { {8}}&{ {7}}&{ {1}} \\ {{ { - 23}}}&{{ {27}}}&{{ { - 69}}} \\ {{ {32}}}&{{ { - 13}}}&{{ {58}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} {{ {24}}}&{{ {12}}}&{ {6}} \\ {{ { - 18}}}&{{ {48}}}&{{ { - 84}}} \\ {{ {42}}}&{{ { - 18}}}&{{ {84}}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} { {5}}&{ {5}}&{ {5}} \\ { {5}}&{{ {10}}}&{{ { - 15}}} \\ {{ {10}}}&{{ { - 5}}}&{{ {15}}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} {{ {11}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {11}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {11}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ {8 - 24 + 5 + 11}}}&{{ {7 - 12 + 5 + 0}}}&{{ {1 - 6 + 5 + 0}}} \\ {{ { - 23 + 18 + 5 + 0}}}&{{ {27 - 48 + 10 + 11}}}&{{ { - 69 + 84 - 15 + 0}}} \\ {{ {32 - 42 + 10 + 0}}}&{{ { - 13 + 18 - 5 + 0}}}&{{ {58 - 84 + 15 + 11}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}}&{ {0}} \\ { {0}}&{ {0}}&{ {0}} \\ { {0}}&{ {0}}&{ {0}} \end{array}} \right]{ { = 0}} \hfill \\ \end{align} \]
= RHS
\[\begin{align} {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A + 11 I = 0}} \hfill \\ {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 5A = - 11 I}} \hfill \\ {{ {A}}^{ {2}}}{ {A}}{{ {A}}^{{ { - 1}}}}{ { - 6AA}}{{ {A}}^{{ { - 1}}}}{ { + 5A}}{{ {A}}^{{ { - 1}}}}{ { = - 11 I }}{{ {A}}^{{ { - 1}}}} \hfill \\ \end{align} \]
\[{ {11 }}{{ {A}}^{{ { - 1}}}}{ { = - }}{{ {A}}^{ {2}}}{ { + 6A - 5I = }}\left[ {\begin{array}{*{20}{c}} {{ { - 4}}}&{{ { - 2}}}&{{ { - 1}}} \\ { {3}}&{{ { - 8}}}&{{ {14}}} \\ {{ { - 7}}}&{ {3}}&{{ { - 14}}} \end{array}} \right]{ { + 6}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {1}}&{ {2}}&{{ { - 3}}} \\ { {2}}&{{ { - 1}}}&{ {3}} \end{array}} \right]{ { - 5}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right]\] \[\begin{align} { { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 4}}}&{{ { - 2}}}&{{ { - 1}}} \\ { {3}}&{{ { - 8}}}&{{ {14}}} \\ {{ { - 7}}}&{ {3}}&{{ { - 14}}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} { {6}}&{ {6}}&{ {6}} \\ { {6}}&{{ {12}}}&{{ { - 18}}} \\ {{ {12}}}&{{ { - 6}}}&{{ {18}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} { {5}}&{ {0}}&{ {0}} \\ { {0}}&{ {5}}&{ {0}} \\ { {0}}&{ {0}}&{ {5}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 3}}}&{ {4}}&{ {5}} \\ { {9}}&{{ { - 1}}}&{{ { - 4}}} \\ { {5}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 3}}}&{ {4}}&{ {5}} \\ { {9}}&{{ { - 1}}}&{{ { - 4}}} \\ { {5}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right] \hfill \\ \end{align} \]
16. यदि \[{ {A}}\;{ { = }}\;\left[ \begin{align} \;2\;\;\;\,\;\; - 1\;\,\;\;\;\;\;1 \hfill \\ - 1\;\;\;\;\;\,\;2\;\;\; - 1 \hfill \\ 1\;\,\;\;\;\,\;\; - 1\;\;\;\;\;\;2 \hfill \\ \end{align} \right]\] तो सत्यापित कीजिए की \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A - 4 I}}\;{ { = }}\;{ {0}}\] है तथा इसकी सहायता से \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए।
उत्तर: \[{ {A}}\;{ { = }}\;\left[ \begin{align} \;2\;\;\;\,\;\; - 1\;\,\;\;\;\;\;1 \hfill \\ - 1\;\;\;\;\;\,\;2\;\;\; - 1 \hfill \\ 1\;\,\;\;\;\,\;\; - 1\;\;\;\;\;\;2 \hfill \\ \end{align} \right]\]
\[\begin{align} {{ {A}}^2}{ { = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{ {1}} \\ {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {1}}&{{ { - 1}}}&{ {2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{ {1}} \\ {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {1}}&{{ { - 1}}}&{ {2}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 5}}}&{ {5}} \\ {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 5}}}&{ {6}} \end{array}} \right] \hfill \\ {{ {A}}^3}{ { = }}{{ {A}}^2}{ {A = }}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 5}}}&{ {5}} \\ {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 5}}}&{ {6}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{ {1}} \\ {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {1}}&{{ { - 1}}}&{ {2}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {22}}}&{{ { - 21}}}&{{ {21}}} \\ {{ { - 21}}}&{{ {22}}}&{{ { - 21}}} \\ {{ {21}}}&{{ { - 21}}}&{{ {22}}} \end{array}} \right] \hfill \\ \end{align} \]
LHS = \[{{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A - 4 I}}\;{ { = }}\;{ {0}}\]
\[{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {22}}}&{{ { - 21}}}&{{ {21}}} \\ {{ { - 21}}}&{{ {22}}}&{{ { - 21}}} \\ {{ {21}}}&{{ { - 21}}}&{{ {22}}} \end{array}} \right]{ { - 6}}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 5}}}&{ {5}} \\ {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 5}}}&{ {6}} \end{array}} \right]{ { + 9}}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{ {1}} \\ {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {1}}&{{ { - 1}}}&{ {2}} \end{array}} \right]{ { - 4}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right]\]
\[{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {22}}}&{{ { - 21}}}&{{ {21}}} \\ {{ { - 21}}}&{{ {22}}}&{{ { - 21}}} \\ {{ {21}}}&{{ { - 21}}}&{{ {22}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} {{ {36}}}&{{ { - 30}}}&{{ {30}}} \\ {{ { - 30}}}&{{ {36}}}&{{ { - 30}}} \\ {{ {30}}}&{{ { - 30}}}&{{ {36}}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} {{ {18}}}&{{ { - 9}}}&{ {9}} \\ {{ { - 9}}}&{{ {18}}}&{{ { - 9}}} \\ { {9}}&{{ { - 9}}}&{{ {18}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{l}} { {4}}&{ {0}}&{ {0}} \\ { {0}}&{ {4}}&{ {0}} \\ { {0}}&{ {0}}&{ {4}} \end{array}} \right]\]
\[\begin{align} { { = }}\left[ {\begin{array}{*{20}{l}} {{ {22 - 36 + 18 - 4}}}&{{ { - 21 + 30 - 9 - 0}}}&{{ { - 21 - 30 + 9 - 0}}} \\ {{ { - 21 + 30 - 9 - 0}}}&{{ {22 - 36 + 18 - 4}}}&{{ { - 21 + 30 - 9 - 0}}} \\ {{ {21 - 30 + 9 - 0}}}&{{ { - 21 + 30 - 9 - 0}}}&{{ {22 - 36 + 18 - 4}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{l}} { {0}}&{ {0}}&{ {0}} \\ { {0}}&{ {0}}&{ {0}} \\ { {0}}&{ {0}}&{ {0}} \end{array}} \right]{ { = 0}} \hfill \\ \end{align} \]
= RHS
\[\begin{align} {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A - 4I = 0}} \hfill \\ {{ {A}}^{ {3}}}{ { - 6}}{{ {A}}^{ {2}}}{ { + 9A = 4I}} \hfill \\ {{ {A}}^{ {2}}}{ {A}}{{ {A}}^{{ { - 1}}}}{ { - 6AA}}{{ {A}}^{{ { - 1}}}}{ { + 9A}}{{ {A}}^{{ { - 1}}}}{ { = 4 I }}{{ {A}}^{{ { - 1}}}} \hfill \\ { {4}}{{ {A}}^{{ { - 1}}}}{ { = }}{{ {A}}^{ {2}}}{ { - 6A + 9I = }}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 5}}}&{ {5}} \\ {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 5}}}&{ {6}} \end{array}} \right]{ { - 6}}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}}&{ {1}} \\ {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {1}}&{{ { - 1}}}&{ {2}} \end{array}} \right]{ { + 9}}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} { { = }}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 5}}}&{ {5}} \\ {{ { - 5}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 5}}}&{ {6}} \end{array}} \right]{ { + }}\left[ {\begin{array}{*{20}{c}} {{ {12}}}&{{ { - 6}}}&{ {6}} \\ {{ { - 6}}}&{{ {12}}}&{{ { - 6}}} \\ { {6}}&{{ { - 6}}}&{{ {12}}} \end{array}} \right]{ { - }}\left[ {\begin{array}{*{20}{c}} { {9}}&{ {0}}&{ {0}} \\ { {0}}&{ {9}}&{ {0}} \\ { {0}}&{ {0}}&{ {9}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}}&{{ { - 1}}} \\ { {1}}&{ {3}}&{ {1}} \\ {{ { - 1}}}&{ {1}}&{ {3}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {1}}&{{ { - 1}}} \\ { {1}}&{ {3}}&{ {1}} \\ {{ { - 1}}}&{ {1}}&{ {3}} \end{array}} \right] \hfill \\ \end{align} \]
17. यदि \[{ {A,}}\;{ {3 \times 3}}\] कोटी का वर्ग आव्यूह है तो \[\left| {{ {adj}}\;{ {A}}} \right|\] का मान है।
(a) \[\left| { {A}} \right|\]
(b) \[{\left| { {A}} \right|^2}\]
(c) \[{\left| { {A}} \right|^3}\]
(d) \[3\left| { {A}} \right|\]
उत्तर: (b) \[{\left| { {A}} \right|^2}\]
18. यदि \[{ {A}}\] कोटी दो का व्युत्क्रमिय आव्यूह है तो \[{ {det(}}{{ {A}}^{{ { - 1}}}}{ {)}}\] बराबर:
(a) \[{ {det(A)}}\]
(b) \[\dfrac{1}{{{ {det(A)}}}}\]
(c) \[1\]
(d) \[0\]
उत्तर: (b) \[\dfrac{1}{{{ {det(A)}}}}\]
प्रश्नावली 4.6
निम्नलिखित प्रश्नों 1 से 6 तक दी गई समीकरण निकायों का संगत अथवा असंगत के वर्गीकरण कीजिए:
1. \[{ {x + 2y}}\;{ { = }}\;{ {2,}}\,\;{ {2x + 3y}}\;{ { = }}\;{ {3}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {2}} \\ { {2}}&{ {3}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {2}} \\ { {3}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {2}} \\ { {2}}&{ {3}} \end{array}} \right|{ { = 1 \times 3 - 2 \times 2 = - 1 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है।
2. \[{ {2x - y}}\;{ { = }}\;5{ {,}}\,\;{ {x + y}}\;{ { = }}\;4\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}} \\ { {1}}&{ {1}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {5}} \\ { {4}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}} \\ { {1}}&{ {1}} \end{array}} \right|{ { = 2 \times 1 - ( - 1) \times 1 = 3 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है।
3. \[{ {x + 3y}}\;{ { = }}\;5{ {,}}\,\;2{ {x + 6y}}\;{ { = }}\;8\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {3}} \\ { {2}}&{ {6}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {5}} \\ { {8}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{l}} { {1}}&{ {3}} \\ { {2}}&{ {6}} \end{array}} \right|{ { = 1 \times 6 - 3 \times 2 = 0}} \hfill \\ \end{align} \]
\[{ {AdjA = }}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 2}}} \\ {{ { - 3}}}&{ {1}} \end{array}} \right]\]
\[{ {(AdjA)B = }}\left[ {\begin{array}{*{20}{c}} { {6}}&{{ { - 2}}} \\ {{ { - 3}}}&{ {1}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {5}} \\ { {8}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} {{ {14}}} \\ {{ { - 7}}} \end{array}} \right]\; \ne \;0\]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व नहीं है। तो यह समीकरण निकाय संगत है।
4. \[{ {x + y + z}}\;{ { = }}\;{ {1,}}\,\;{ {2x + 3y + 2z}}\;{ { = }}\;{ {2,}}\;{ {ax + ay + 2az}}\;{ { = }}\;{ {4}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {2}}&{ {3}}&{ {2}} \\ { {a}}&{ {a}}&{{ {2a}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {1}} \\ { {2}} \\ { {4}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {2}}&{ {3}}&{ {2}} \\ { {a}}&{ {a}}&{{ {2a}}} \end{array}} \right|{ { = 1(3 \times 2a - 2 \times a) - 1(2 \times 2a - 2 \times a) + (2 \times a - 3 \times a)}} \hfill \\ { { = a }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है।
5. \[{ {3x - y - 2z}}\;{ { = }}\;{ {2,}}\,\;{ {2y - z}}\;{ { = }}\;{ { - 1,}}\;{ {3x - 5y}}\;{ { = }}\;{ {3}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}}&{{ { - 2}}} \\ { {0}}&{ {2}}&{{ { - 1}}} \\ { {3}}&{{ { - 5}}}&{ {0}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{c}} { {2}} \\ {{ { - 1}}} \\ { {3}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}}&{{ { - 2}}} \\ { {0}}&{ {2}}&{{ { - 1}}} \\ { {3}}&{{ { - 5}}}&{ {0}} \end{array}} \right|{ { = - 15 + 3 + 12 = 0}} \hfill \\ { {Adj A = }}\left[ {\begin{array}{*{20}{c}} {{ { - 5}}}&{{ {10}}}&{ {5}} \\ {{ { - 3}}}&{ {6}}&{ {3}} \\ {{ { - 6}}}&{{ {12}}}&{ {6}} \end{array}} \right] \hfill \\ { {(AdjA)B = }}\left[ {\begin{array}{*{20}{c}} {{ { - 5}}}&{{ {10}}}&{ {5}} \\ {{ { - 3}}}&{ {6}}&{ {3}} \\ {{ { - 6}}}&{{ {12}}}&{ {6}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {2}} \\ {{ { - 1}}} \\ { {3}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 10 - 10 + 15}}} \\ {{ { - 6 - 6 + 9}}} \\ {{ { - 12 - 12 + 18}}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ { = }}\left[ {\begin{array}{*{20}{l}} {{ { - 5}}} \\ {{ { - 3}}} \\ {{ { - 6}}} \end{array}} \right]{ { }} \ne { { 0}}\]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व नहीं है। तो यह समीकरण निकाय असंगत है।
6. \[{ {5x - y + 4z}}\;{ { = }}\;{ {5,}}\,\;{ {2x + 3y - 5z}}\;{ { = }}\;{ {2,}}\;{ {5x - 2y + 6z}}\;{ { = }}\;{ { - 1}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {5}}&{{ { - 1}}}&{ {4}} \\ { {2}}&{ {3}}&{ {5}} \\ { {5}}&{{ { - 2}}}&{ {6}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{c}} { {5}} \\ { {2}} \\ {{ { - 1}}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {5}}&{{ { - 1}}}&{ {4}} \\ { {2}}&{ {3}}&{ {5}} \\ { {5}}&{{ { - 2}}}&{ {6}} \end{array}} \right|{ { = 5(18 + 3) - 1(12 - 25) + 4( - 4 - 15)}} \hfill \\ { { = 5 \times 28 - 13 - 4 \times 5 = 67 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है।
निम्नलिखित प्रश्न 7 से 17 तक प्रत्येक समीकरण निकाय को आव्यूह विधि से हल कीजिए:
7. \[{ {5x + 2y}}\;{ { = }}\;4{ {,}}\,\;7{ {x + 3y}}\;{ { = }}\;5\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{l}} { {5}}&{ {2}} \\ { {7}}&{ {3}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {4}} \\ { {5}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{l}} { {5}}&{ {2}} \\ { {7}}&{ {3}} \end{array}} \right|{ { = 5 \times 3 - 7 \times 2 = 15 - 14 = 1 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {Adj A = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{{ { - 2}}} \\ {{ { - 7}}}&{ {5}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{{ { - 2}}} \\ {{ { - 7}}}&{ {5}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ {X = }}{{ {A}}^{{ { - 1}}}}{ {B = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{{ { - 2}}} \\ {{ { - 7}}}&{ {5}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {4}} \\ { {5}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {(12 - 10)}}} \\ {{ {( - 28 + 25)}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {2}} \\ {{ { - 3}}} \end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {2}} \\ {{ { - 3}}} \end{array}} \right]\]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = 2, y = - 3}}\]
8. \[{ {2x - y}}\;{ { = }}\; - 2{ {,}}\,\;3{ {x + 4y}}\;{ { = }}\;3\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}} \\ { {3}}&{ {4}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{c}} {{ { - 2}}} \\ { {3}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {2}}&{{ { - 1}}} \\ { {3}}&{ {4}} \end{array}} \right|{ { = 2 \times 4 + 1 \times 3 = 8 + 3 = 11 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {Adj A = }}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {1}} \\ {{ { - 3}}}&{ {2}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {1}} \\ {{ { - 3}}}&{ {2}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} { {X = }}{{ {A}}^{{ { - 1}}}}{ {B = }}\dfrac{{ {1}}}{{{ {11}}}} \hfill \\ \left[ {\begin{array}{*{20}{c}} { {4}}&{ {1}} \\ {{ { - 3}}}&{ {2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{ { - 2}}} \\ { {3}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {(4 \times - 2 + 1 \times 3)}}} \\ {{ {( - 3 \times - 2 + 2 \times 3)}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 5}}} \\ {{ {12}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {\dfrac{{{ { - 5}}}}{{{ {11}}}}} \\ {\dfrac{{{ {12}}}}{{{ {11}}}}} \end{array}} \right] \hfill \\ \end{align} \]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = - }}\dfrac{{ {5}}}{{{ {11}}}}{ {, y = }}\dfrac{{{ {12}}}}{{{ {11}}}}\]
9. \[{ {4x - 3y}}\;{ { = }}\;3{ {,}}\,\;3{ {x - 5y}}\;{ { = }}\;7\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{l}} { {4}}&{{ { - 3}}} \\ { {3}}&{{ { - 5}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {3}} \\ { {7}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{l}} { {4}}&{{ { - 3}}} \\ { {3}}&{{ { - 5}}} \end{array}} \right|{ { = 4 \times - 5 + 3 \times 3 = - 20 + 9 = - 11 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {Adj A = }}\left[ {\begin{array}{*{20}{l}} {{ { - 5}}}&{ {3}} \\ {{ { - 3}}}&{ {4}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = - }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{l}} {{ { - 5}}}&{ {3}} \\ {{ { - 3}}}&{ {4}} \end{array}} \right] \hfill \\ { {X = }}{{ {A}}^{{ { - 1}}}}{ {B = }}\dfrac{{ {1}}}{{{ {11}}}} \hfill \\ \left[ {\begin{array}{*{20}{l}} {{ { - 5}}}&{ {3}} \\ {{ { - 3}}}&{ {4}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {3}} \\ { {7}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} {{ {( - 5 \times 3 + 3 \times 7)}}} \\ {{ {( - 3 \times 3 + 4 \times 7)}}} \end{array}} \right]{ { = - }}\dfrac{{ {1}}}{{{ {11}}}}\left[ {\begin{array}{*{20}{c}} { {6}} \\ {{ {19}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} {{ { - }}\dfrac{{ {6}}}{{{ {11}}}}} \\ {{ { - }}\dfrac{{{ {19}}}}{{{ {11}}}}} \end{array}} \right] \hfill \\ \end{align} \]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = }}\dfrac{{ {6}}}{{{ {11}}}}{ {, y = - }}\dfrac{{{ {19}}}}{{{ {11}}}}\]
10. \[{ {5x + 2y}}\;{ { = }}\;3{ {,}}\,\;3{ {x + 2y}}\;{ { = }}\;5\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{l}} { {5}}&{ {2}} \\ { {3}}&{ {2}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {3}} \\ { {7}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{l}} { {5}}&{ {2}} \\ { {3}}&{ {2}} \end{array}} \right|{ { = 5 \times 2 - 3 \times 2}} \hfill \\ \end{align} \]
\[{ { = 10 - 6 = 4 }} \ne { { 0}}\]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[{ {Adj A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 2}}} \\ {{ { - 3}}}&{ {5}} \end{array}} \right]\]
\[{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 2}}} \\ {{ { - 3}}}&{ {5}} \end{array}} \right]\]
\[\begin{align} { {X = }}{{ {A}}^{{ { - 1}}}}{ {B = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 2}}} \\ {{ { - 3}}}&{ {5}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {3}} \\ { {5}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {(2 \times 3 - 2 \times 5)}}} \\ {{ {( - 3 \times 3 + 5 \times 5)}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} {{ { - 4}}} \\ {{ {16}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}} \\ {{ {16}}} \end{array}} \right] \hfill \\ \end{align} \]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = - 1, y = 4}}\]
11. \[{ {2x + y + z}}\;{ { = }}\;{ {1,}}\,\;{ {x - 2y - z}}\;{ { = }}\;\dfrac{{ {3}}}{{ {2}}}{ {,}}\;{ {3y - 5z}}\;{ { = }}\;{ {9}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {1}}&{ {1}} \\ { {1}}&{{ { - 2}}}&{{ { - 1}}} \\ { {0}}&{ {3}}&{{ { - 5}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {1}} \\ {\dfrac{{ {3}}}{{ {2}}}} \\ { {9}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {2}}&{ {1}}&{ {1}} \\ { {1}}&{{ { - 2}}}&{{ { - 1}}} \\ { {0}}&{ {3}}&{{ { - 5}}} \end{array}} \right|{ { = 2(10 + 3) - 1( - 5 - 3) + 0( - 1 + 3)}} \hfill \\ { { = 26 + 8 + 0 = 34 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {AdjA = }}\left[ {\begin{array}{*{20}{c}} {{ {13}}}&{ {8}}&{ {1}} \\ { {5}}&{{ { - 10}}}&{ {3}} \\ { {3}}&{{ { - 6}}}&{{ { - 5}}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}} {{ {13}}}&{ {8}}&{ {1}} \\ { {5}}&{{ { - 10}}}&{ {3}} \\ { {3}}&{{ { - 6}}}&{{ { - 5}}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ {X = }}{{ {A}}^{{ { - 1}}}}{ {B}}\]
\[{ { = }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}} {{ {13}}}&{ {8}}&{ {1}} \\ { {5}}&{{ { - 10}}}&{ {3}} \\ { {3}}&{{ { - 6}}}&{{ { - 5}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {1}} \\ {\dfrac{{ {3}}}{{ {2}}}} \\ { {9}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}} {\left( {{ {13 \times 1 + 8 \times }}\dfrac{{ {3}}}{{ {2}}}{ { + 1 \times 9}}} \right)} \\ {\left( {{ {5 \times 1 - 10 \times }}\dfrac{{ {3}}}{{ {2}}}{ { + 3 \times 9}}} \right)} \\ {\left( {{ {3 \times 1 - 6 \times }}\dfrac{{ {3}}}{{ {2}}}{ { - 5 \times 9}}} \right)} \end{array}} \right]\]
\[{ { = }}\dfrac{{ {1}}}{{{ {34}}}}\left[ {\begin{array}{*{20}{c}} {{ {34}}} \\ {{ {17}}} \\ {{ { - 51}}} \end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {1}} \\ {\dfrac{{ {1}}}{{ {2}}}} \\ {{ { - }}\dfrac{{ {3}}}{{ {2}}}} \end{array}} \right]\]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = 1, y = }}\dfrac{{ {1}}}{{ {2}}}{ {, Z = - }}\dfrac{{ {3}}}{{ {2}}}\]
12. \[{ {x - y + z}}\;{ { = }}\;{ {4,}}\,\;{ {2x + y - 3z}}\;{ { = }}\;{ {0,}}\;{ {x + y + z}}\;{ { = }}\;{ {2}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {1}} \\ { {2}}&{ {1}}&{{ { - 3}}} \\ { {1}}&{ {1}}&{ {1}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {4}} \\ { {0}} \\ { {2}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {1}} \\ { {2}}&{ {1}}&{{ { - 3}}} \\ {{ {11}}}&{ {1}}&{ {1}} \end{array}} \right|{ { = 1(1 + 3) + 1(2 + 3) + 1(2 - 1)}} \hfill \\ { { = 4 + 5 + 1 = 10 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {Adj A = }}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {2}}&{ {2}} \\ {{ { - 5}}}&{ {0}}&{ {5}} \\ { {1}}&{{ { - 2}}}&{ {3}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {2}}&{ {2}} \\ {{ { - 5}}}&{ {0}}&{ {5}} \\ { {1}}&{{ { - 2}}}&{ {3}} \end{array}} \right] \hfill \\ { {X = }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\ { { = }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}} { {4}}&{ {2}}&{ {2}} \\ {{ { - 5}}}&{ {0}}&{ {5}} \\ { {1}}&{{ { - 2}}}&{ {3}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {4}} \\ { {0}} \\ { {2}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}} {{ {(4 \times 4 + 2 \times 0 + 2 \times 2)}}} \\ {{ {( - 5 \times 4 + 0 \times 0 + 5 \times 2)}}} \\ {{ {(1 \times 4 - 2 \times 0 - 3 \times 2)}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}} {{ {16 + 0 + 4}}} \\ {{ { - 20 + 0 + 10}}} \\ {{ {4 + 0 + 6}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {10}}}}\left[ {\begin{array}{*{20}{c}} {{ {20}}} \\ {{ { - 10}}} \\ {{ {10}}} \end{array}} \right] \hfill \\ \end{align} \]
\[\left[ {\begin{array}{*{20}{c}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {2}} \\ {{ { - 1}}} \\ { {1}} \end{array}} \right]\]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = 2, y = - 1, Z = 1}}\]
13. \[{ {2x + 3y + 3z}}\;{ { = }}\;5{ {,}}\,\;{ {x - 2y + z}}\;{ { = }}\; - 4{ {,}}\;3{ {x - y - 2z}}\;{ { = }}\;3\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {3}}&{ {3}} \\ { {1}}&{{ { - 2}}}&{ {1}} \\ { {3}}&{{ { - 1}}}&{{ { - 2}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{c}} { {5}} \\ {{ { - 4}}} \\ { {3}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {2}}&{ {3}}&{ {3}} \\ { {1}}&{{ { - 2}}}&{ {1}} \\ { {3}}&{{ { - 1}}}&{{ { - 2}}} \end{array}} \right|{ { = 2(4 + 1) - 3( - 2 - 3) + 3( - 1 + 6) = 10 + 15 + 15 = 40 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {AdjA = }}\left[ {\begin{array}{*{20}{c}} { {5}}&{ {3}}&{ {9}} \\ { {5}}&{{ { - 13}}}&{ {1}} \\ { {5}}&{{ {11}}}&{{ { - 7}}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}} { {5}}&{ {3}}&{ {9}} \\ { {5}}&{{ { - 13}}}&{ {1}} \\ { {5}}&{{ {11}}}&{{ { - 7}}} \end{array}} \right] \hfill \\ { {X = }}{{ {A}}^{{ { - 1}}}}{ {B = }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}} { {5}}&{ {3}}&{ {9}} \\ { {5}}&{{ { - 13}}}&{ {1}} \\ { {5}}&{{ {11}}}&{{ { - 7}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {5}} \\ {{ { - 4}}} \\ { {3}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}} {{ {(5 \times 5 + 3 \times - 4 + 9 \times 3)}}} \\ {{ {(5 \times 5 - 13 \times - 4 + 1 \times 3)}}} \\ {{ {(5 \times 5 + 11 \times - 4 - 7 \times 3)}}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}} {{ {25 - 12 + 27}}} \\ {{ {25 + 52 + 3}}} \\ {{ {25 - 44 - 21}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {40}}}}\left[ {\begin{array}{*{20}{c}} {{ {40}}} \\ {{ {80}}} \\ {{ { - 40}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {1}} \\ { {2}} \\ {{ { - 1}}} \end{array}} \right] \hfill \\ \end{align} \]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = 1, y = 2, Z = - 1}}\]
14. \[{ {x - y + 2z}}\;{ { = }}\;{ {7,}}\,\;{ {3x + 4y - 5z}}\;{ { = }}\;{ { - 5,}}\;{ {2x - y + 3z}}\;{ { = }}\;{ {12}}\]
उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {3}}&{ {4}}&{{ { - 5}}} \\ { {2}}&{{ { - 1}}}&{{ { - 3}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{c}} { {7}} \\ {{ { - 5}}} \\ {{ {12}}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{c}} { {1}}&{{ { - 1}}}&{ {2}} \\ { {3}}&{ {4}}&{{ { - 5}}} \\ { {2}}&{{ { - 1}}}&{{ { - 3}}} \end{array}} \right| \hfill \\ { { = 1(12 - 5) + 1(9 + 10) + 2( - 3 - 8)}} \hfill \\ { { = 7 + 19 - 22 = 4 }} \ne { { 0}} \hfill \\ \end{align} \]
इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।
\[\begin{align} { {Adj A = }}\left[ {\begin{array}{*{20}{c}} { {7}}&{ {1}}&{{ { - 3}}} \\ {{ { - 19}}}&{{ { - 1}}}&{{ {11}}} \\ {{ { - 11}}}&{{ { - 1}}}&{ {7}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} { {7}}&{ {1}}&{{ { - 3}}} \\ {{ { - 19}}}&{{ { - 1}}}&{{ {11}}} \\ {{ { - 11}}}&{{ { - 1}}}&{ {7}} \end{array}} \right] \hfill \\ { {X = }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} { {7}}&{ {1}}&{{ { - 3}}} \\ {{ { - 19}}}&{{ { - 1}}}&{{ {11}}} \\ {{ { - 11}}}&{{ { - 1}}}&{ {7}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { {7}} \\ {{ { - 5}}} \\ {{ {12}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} {{ {(7 \times 7 + 1 \times - 5 - 3 \times 12)}}} \\ {{ {( - 19 \times 7 - 1 \times - 5 + 11 \times 12)}}} \\ {{ {( - 11 \times 7 - 1 \times - 5 + 7 \times 12)}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} {{ {49 - 5 - 36}}} \\ {{ { - 133 + 5 + 132}}} \\ {{ { - 77 + 5 + 84}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{ {4}}}\left[ {\begin{array}{*{20}{c}} { {8}} \\ { {4}} \\ {{ {12}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{c}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {2}} \\ { {1}} \\ { {3}} \end{array}} \right] \hfill \\ \end{align} \]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = 2, y = 1, Z = 3}}\]
15. यदि \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 3}}}&{ {5}} \\ { {3}}&{ {2}}&{{ { - 4}}} \\ { {1}}&{ {1}}&{{ { - 2}}} \end{array}} \right]\] है तो \[{{ {A}}^{{ { - 1}}}}\] ज्ञात कीजिए, \[{{ {A}}^{{ { - 1}}}}\] का प्रयोग करके निम्नलिखित समीकरण निकाय को हल कीजिए:
\[{ {2x - 3y + 5z = 11 , 3x + 2y - 4z = - 5 , x + y - 2z = - 3}}\]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{{ { - 3}}}&{ {5}} \\ { {3}}&{ {2}}&{{ { - 4}}} \\ { {1}}&{ {1}}&{{ { - 2}}} \end{array}} \right]\]
\[\begin{align} { {|A| = }}\left| {\begin{array}{*{20}{c}} { {2}}&{{ { - 3}}}&{ {5}} \\ { {3}}&{ {2}}&{{ { - 4}}} \\ { {1}}&{ {1}}&{{ { - 2}}} \end{array}} \right|{ { = 2( - 4 + 4) + 3( - 6 + 4) + 5(3 - 2)}} \hfill \\ { { = 0 - 6 + 5 = - 1}} \hfill \\ { {AdjA = }}\left[ {\begin{array}{*{20}{c}} { {0}}&{{ { - 1}}}&{ {2}} \\ { {2}}&{{ { - 9}}}&{{ {23}}} \\ { {1}}&{{ { - 5}}}&{{ {13}}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = - }}\left[ {\begin{array}{*{20}{l}} { {0}}&{{ { - 1}}}&{ {2}} \\ { {2}}&{{ { - 9}}}&{{ {23}}} \\ { {1}}&{{ { - 5}}}&{{ {13}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{r}} {{ { - 0}}}&{ {1}}&{{ { - 2}}} \\ {{ { - 2}}}&{ {9}}&{{ { - 23}}} \\ {{ { - 1}}}&{ {5}}&{{ { - 13}}} \end{array}} \right] \hfill \\ \because \;{ {X = }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\ \therefore \;{ {B = }}\left[ {\begin{array}{*{20}{c}} {{ {11}}} \\ {{ { - 5}}} \\ {{ { - 3}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right] \hfill \\ { {2x - 3y + 5z = 11, 3x + 2y - 4z = - 5, x + y - 2z = - 3}} \hfill \\ \end{align} \]
\[\begin{align} { {X = }}\left[ {\begin{array}{*{20}{l}} {{ { - 0}}}&{ {1}}&{{ { - 2}}} \\ {{ { - 2}}}&{ {9}}&{{ { - 23}}} \\ {{ { - 1}}}&{ {5}}&{{ { - 13}}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{ {11}}} \\ {{ { - 5}}} \\ {{ { - 3}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{l}} {{ {( - 0 \times 11 - 1 \times 5 + 2 \times 3)}}} \\ {{ {( - 2 \times 11 - 9 \times 5 + 23 \times 3)}}} \\ {{ {( - 1 \times 11 - 5 \times 5 + 13 \times 3)}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ {0 - 5 + 6}}} \\ {{ { - 22 - 45 + 69}}} \\ {{ { - 11 - 25 + 39}}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ { = }}\left[ {\begin{array}{*{20}{l}} { {1}} \\ { {2}} \\ { {3}} \end{array}} \right]\]
\[\left[ {\begin{array}{*{20}{l}} { {y}} \\ { {z}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {1}} \\ { {2}} \\ { {3}} \end{array}} \right]\]
इसलिए, इस समीकरण निकाय का हल है
\[{ {x = 1, y = 2, Z = 3}}\]
16. \[{ {4\;kg}}\] प्याज, \[{ {3\;kg}}\] गेहू और \[{ {2\;kg}}\] चावल का मूल्य \[{ {Rs}}{ {. 60}}\] है। \[{ {2\;kg}}\] प्याज, \[{ {4\;kg}}\] गेहू और \[{ {6\;kg}}\] चावल का मूल्य \[{ {Rs}}{ {. 90}}\] है। \[{ {6\;kg}}\] प्याज, \[{ {2\;kg}}\] गेहू और \[{ {3 kg}}\] चावल का मूल्य \[{ {Rs}}{ {. 70}}\] है। आव्यूह विधि द्वारा प्रत्येक का मूल्य प्रति \[{ {kg}}\] ज्ञात कीजिए।
उत्तर: मान लेते है की प्याज का मूल्य प्रति \[{ {kg Rs}}{ {. x}}\] , गेहू का मूल्य प्रति \[{ {kg Rs}}{ {. y}}\] , तथा चवक का मूल्य प्रति \[{ {kg Rs}}{ {. z}}\] है।
इसलिए हम फी गई जानकारी को समीकरणों के रूप मे कुछ इस प्रकार रख सकते है:
\[{ {4x + 3y + 2z = 60}}\]
\[\begin{align} { {2x + 4y + 6z = 90}} \hfill \\ { {6x + 2y + 3y = 70}} \hfill \\ \end{align} \]
दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है।
\[\begin{align} { {AX = B}} \hfill \\ { {A = }}\left[ {\begin{array}{*{20}{l}} { {4}}&{ {3}}&{ {2}} \\ { {2}}&{ {4}}&{ {6}} \\ { {6}}&{ {2}}&{ {3}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} {{ {60}}} \\ {{ {90}}} \\ {{ {70}}} \end{array}} \right] \hfill \\ { {|A| = }}\left| {\begin{array}{*{20}{l}} { {4}}&{ {3}}&{ {2}} \\ { {2}}&{ {4}}&{ {6}} \\ { {6}}&{ {2}}&{ {3}} \end{array}} \right|{ { = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 0 + 90 - 40 = 50 }} \ne { { 0}} \hfill \\ { {AdjA = }}\left[ {\begin{array}{*{20}{c}} { {0}}&{{ { - 5}}}&{{ {10}}} \\ {{ {30}}}&{ {0}}&{{ { - 20}}} \\ {{ { - 20}}}&{{ {10}}}&{{ {10}}} \end{array}} \right] \hfill \\ {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{{ {AdjA}}}}{{{ {|A|}}}}{ { = }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}} { {0}}&{{ { - 5}}}&{{ {10}}} \\ {{ {30}}}&{ {0}}&{{ { - 20}}} \\ {{ { - 20}}}&{{ {10}}}&{{ {10}}} \end{array}} \right] \hfill \\ { {X = }}{{ {A}}^{{ { - 1}}}}{ {B = }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}} { {0}}&{{ { - 5}}}&{{ {10}}} \\ {{ {30}}}&{ {0}}&{{ { - 20}}} \\ {{ { - 20}}}&{{ {10}}}&{{ {10}}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{ {60}}} \\ {{ {90}}} \\ {{ {70}}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ { = }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}} {{ {(0 \times 60 - 5 \times 90 + 10 \times 70)}}} \\ {{ {(30 \times 60 + 0 \times 90 - 20 \times 70)}}} \\ {{ {( - 20 \times 60 + 10 \times 90 + 10 \times 70)}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{c}} {{ {0 - 450 + 700}}} \\ {{ {1800 + 0 - 1400}}} \\ {{ { - 1200 + 900 + 700}}} \end{array}} \right]\]
\[\begin{align} { { = }}\dfrac{{ {1}}}{{{ {50}}}}\left[ {\begin{array}{*{20}{l}} {{ {250}}} \\ {{ {400}}} \\ {{ {400}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{l}} { {x}} \\ { {y}} \\ { {z}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} { {5}} \\ { {8}} \\ { {8}} \end{array}} \right] \hfill \\ \end{align} \]
इसलिए इस समीकरण निकाय का हल है:
\[{ {x = 5, y = 8, Z = 8}}\]
प्याज का मूल्य प्रति \[{ {kg Rs}}{ {. 5}}\] , गेहू का मूल्य प्रति \[{ {kg Rs}}{ {. 8}}\] , तथा चावल का मूल्य प्रति \[{ {kg Rs}}{ {. 8}}\] है।
प्रश्नावली A4
1. सिद्ध कीजिए की सारणिक \[\left| {\begin{array}{*{20}{c}} { {x}}&{{ {sin\theta }}}&{{ {cos\theta }}} \\ {{ { - sin\theta }}}&{{ { - x}}}&{ {1}} \\ {{ {cos\theta }}}&{ {1}}&{ {x}} \end{array}} \right|{ { \theta }}\] से स्वतंत्र है।
उत्तर: \[{ {\Delta = }}\left| {\begin{array}{*{20}{c}} { {x}}&{{ {sin\theta }}}&{{ {cos\theta }}} \\ {{ { - sin\theta }}}&{{ { - x}}}&{ {1}} \\ {{ {cos\theta }}}&{ {1}}&{ {x}} \end{array}} \right|\]
\[\begin{align} { { = x - }}\left( {{{ {x}}^{ {2}}}{ { - 1}}} \right){ { - sin\theta ( - xsin\theta - cos\theta ) + cos\theta ( - sin\theta + xcos\theta )}} \hfill \\ { { = - }}{{ {x}}^{ {3}}}{ { - x + xsi}}{{ {n}}^{ {2}}}{ {\theta + sin\theta cos\theta - cos\theta sin\theta + xco}}{{ {s}}^{ {2}}}{ {\theta }} \hfill \\ { { = - }}{{ {x}}^{ {3}}}{ { - x + x}}\left( {{ {si}}{{ {n}}^{ {2}}}{ {\theta + co}}{{ {s}}^{ {2}}}{ {\theta }}} \right) \hfill \\ { { = - }}{{ {x}}^{ {3}}}{ { - x + x}} \hfill \\ { { = - }}{{ {x}}^{ {2}}} \hfill \\ \end{align} \]
जो \[{ {\theta }}\] से स्वतंत्र है।
2. सारणिक का प्रसरण किए बिना सिद्ध कीजिए की \[\left| {\begin{array}{*{20}{l}} { {a}}&{{{ {a}}^{ {2}}}}&{{ {bc}}} \\ { {b}}&{{{ {b}}^{ {2}}}}&{{ {bc}}} \\ { {c}}&{{{ {c}}^{ {2}}}}&{{ {ab}}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}} \\ { {1}}&{{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}} \\ { {1}}&{{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|\]
उत्तर: \[\left| {\begin{array}{*{20}{l}} { {a}}&{{{ {a}}^{ {2}}}}&{{ {bc}}} \\ { {b}}&{{{ {b}}^{ {2}}}}&{{ {ca}}} \\ { {c}}&{{{ {c}}^{ {2}}}}&{{ {ab}}} \end{array}} \right|{ { = }}\dfrac{{ {1}}}{{{ {abc}}}}\left| {\begin{array}{*{20}{l}} {{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}}&{{ {abc}}} \\ {{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}}&{{ {abc}}} \\ {{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}&{{ {abc}}} \end{array}} \right|\quad \;\;\;\;\;{ {[}}{{ {R}}_1} \to { {a}}{{ {R}}_1}{ {, }}{{ {R}}_2} \to { {b}}{{ {R}}_2}{ {, }}{{ {R}}_3} \to { {c}}{{ {R}}_3}{ {]}}\]
\[{ { = }}\dfrac{{{ {abc}}}}{{{ {abc}}}}\left| {\begin{array}{*{20}{l}} {{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}}&{ {1}} \\ {{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}}&{ {1}} \\ {{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}}&{ {1}} \end{array}} \right|\quad \] [\[{{ {C}}_3}\] से \[{ {abc}}\] आम लेना]
\[\begin{align} { { = ( - 1}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{l}} { {1}}&{{{ {a}}^{ {3}}}}&{{{ {a}}^{ {2}}}} \\ { {1}}&{{{ {b}}^{ {3}}}}&{{{ {b}}^{ {2}}}} \\ { {1}}&{{{ {c}}^{ {3}}}}&{{{ {c}}^{ {2}}}} \end{array}} \right|\,\;\;\,{ { }}\,\,\,{ { [}}{{ {C}}_1} \leftrightarrow {{ {C}}_3}{ {]}} \hfill \\ { { = ( - 1}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{l}} { {1}}&{{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}} \\ { {1}}&{{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}} \\ { {1}}&{{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|\quad \;\;\;\;\;\;{ {[}}{{ {C}}_2} \leftrightarrow {{ {C}}_3}{ {]}} \hfill \\ \end{align} \]
\[{ { = }}\left| {\begin{array}{*{20}{l}} { {1}}&{{{ {a}}^{ {2}}}}&{{{ {a}}^{ {3}}}} \\ { {1}}&{{{ {b}}^{ {2}}}}&{{{ {b}}^{ {3}}}} \\ { {1}}&{{{ {c}}^{ {2}}}}&{{{ {c}}^{ {3}}}} \end{array}} \right|\]
3. \[\left| {\begin{array}{*{20}{c}} {{ {cos\alpha cos\beta }}}&{{ {cos\alpha sin\beta }}}&{{ { - sin\alpha }}} \\ {{ { - sin\beta }}}&{{ {cos\beta }}}&{ {0}} \\ {{ {sin\alpha cos\beta }}}&{{ {sin\alpha sin\beta }}}&{{ {cos\alpha }}} \end{array}} \right|\] का मान ज्ञात कीजिए।
उत्तर: \[{ { = - sin\alpha }}\left( {{ { - sin\alpha si}}{{ {n}}^{ {2}}}{ {\beta - sin\alpha co}}{{ {s}}^{ {2}}}{ {\beta }}} \right){ { - 0(cos\alpha cos\beta sin\alpha sin\beta - cos\alpha sin\beta sin\alpha cos\beta )}}\]
\[{ { + cos\alpha }}\left( {{ {cos\alpha co}}{{ {s}}^{ {2}}}{ {\beta + cos\alpha co}}{{ {s}}^{ {2}}}{ {\beta }}} \right)\]
\[\begin{align} { { = si}}{{ {n}}^{ {2}}}{ {\alpha }}\left( {{ {si}}{{ {n}}^{ {2}}}{ {\beta + co}}{{ {s}}^{ {2}}}{ {\beta }}} \right){ { + co}}{{ {s}}^{ {2}}}{ {\alpha }}\left( {{ {co}}{{ {s}}^{ {2}}}{ {\beta + si}}{{ {n}}^{ {2}}}{ {\beta }}} \right) \hfill \\ { { = si}}{{ {n}}^{ {2}}}{ {\alpha + co}}{{ {s}}^{ {2}}}{ {\alpha }} \hfill \\ { { = 1}} \hfill \\ \end{align} \]
4. यदि \[{ {a, b, c}}\] वास्तविक संखयाए हो और सारणिक \[{ {\Delta = }}\left| {\begin{array}{*{20}{l}} {{ {b + c}}}&{{ {c + a}}}&{{ {a + b}}} \\ {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\ {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}} \end{array}} \right|{ { = 0}}\] हो तो दर्शाइए की या तो \[{ {a + b + c = 0}}\] या \[{ {a = b = c}}\] है।
उत्तर: \[\left| {\begin{array}{*{20}{l}} {{ {b + c}}}&{{ {c + a}}}&{{ {a + b}}} \\ {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\ {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}} \end{array}} \right|{ { = 0}}\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {2(a + b + c)}}}&{{ {2(a + b + c)}}}&{{ {2(a + b + c)}}} \\ {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\ {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}} \end{array}} \right|{ { = 0}}\;\,\quad { {[}}{{ {R}}_1} \to {{ {R}}_1}{ { + }}{{ {R}}_2}{ { + }}{{ {R}}_3}{ {]}}\]
\[{ { = 2(a + b + c)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\ {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}} \end{array}} \right|{ { = 0}}\]
[\[{{ {R}}_1}\] के सामान्य रूप से \[{ {2(a + b + c)}}\] ]
\[{ { = 2(a + b + c)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {0}}&{ {0}} \\ {{ {c + a}}}&{{ {a + b}}}&{{ {b + c}}} \\ {{ {a + b}}}&{{ {b + c}}}&{{ {c + a}}} \end{array}} \right|{ { = 0}}\;\;\,\,\;\;\;{ { [}}{{ {C}}_2} \to {{ {C}}_2}{ { - C1, }}{{ {C}}_3} \to {{ {C}}_3}{ { - }}{{ {C}}_1}]\]
\[\begin{align} \therefore \;{ {2(a + b + c)[(b - c)(c - b) - (b - a)(c - a)] = 0}} \hfill \\ { {2(a + b + c)}}\left[ {{ {bc - }}{{ {b}}^{ {2}}}{ { - }}{{ {c}}^{ {2}}}{ { + bc - }}\left( {{ {bc - ab - ac + }}{{ {a}}^{ {2}}}} \right)} \right]{ { = 0}} \hfill \\ \end{align} \]
\[\begin{align} { {2(a + b + c)}}\left[ {{ {bc - }}{{ {b}}^{ {2}}}{ { - }}{{ {c}}^{ {2}}}{ { + bc - bc + ab + ac - }}{{ {a}}^{ {2}}}} \right]{ { = 0}} \hfill \\ { { - (a + b + c)}}\left[ {{ {2}}{{ {a}}^{ {2}}}{ { + 2}}{{ {b}}^{ {2}}}{ { + 2}}{{ {c}}^{ {2}}}{ { - 2ab - 2bc - 2ca}}} \right]{ { = 0}} \hfill \\ \end{align} \]
\[\begin{align} { { - (a + b + c)}}\left[ {\left( {{{ {a}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + 2ab}}} \right){ { + }}\left( {{{ {b}}^{ {2}}}{ { + }}{{ {c}}^{ {2}}}{ { - 2bc}}} \right)} \right.\left. {{ { + }}\left( {{{ {c}}^{ {2}}}{ { + }}{{ {a}}^{ {2}}}{ { - 2ca}}} \right)} \right]{ { = 0}} \hfill \\ { { - (a + b + c)}}\left[ {{{{ {(a - b)}}}^{ {2}}}{ { + (b - c}}{{ {)}}^{ {2}}}{ { + (c - a}}{{ {)}}^{ {2}}}} \right]{ { = 0}} \hfill \\ \end{align} \]
\[{ {a + b + c = 0 ; (a - b}}{{ {)}}^{ {2}}}{ { = 0, (b - c}}{{ {)}}^{ {2}}}{ { = 0, (c - a}}{{ {)}}^{ {2}}}{ { = 0}}\]
\[\begin{align} { {a + b + c = 0 ; (a - b) = 0, (b - c) = 0, (c - a) = 0}} \hfill \\ { {a + b + c = 0 ; a = b, b = c, c = a}} \hfill \\ { {a + b + c = 0 ; a = b = c}} \hfill \\ \end{align} \]
5. यदि \[{ {a }} \ne { { 0}}\] हो तो समीकरण \[\left| {\begin{array}{*{20}{c}} {{ {x + a}}}&{ {x}}&{ {x}} \\ { {x}}&{{ {x + a}}}&{ {x}} \\ { {x}}&{ {x}}&{{ {x + a}}} \end{array}} \right|{ { = 0}}\] को हल कीजिए।
उत्तर: \[\left| {\begin{array}{*{20}{c}} {{ {x + a}}}&{ {x}}&{ {x}} \\ { {x}}&{{ {x + a}}}&{ {x}} \\ { {x}}&{ {x}}&{{ {x + a}}} \end{array}} \right|{ { = 0}}\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {3x + a}}}&{{ {3x + a}}}&{{ {3x + a}}} \\ { {x}}&{{ {x + 2}}}&{ {x}} \\ { {x}}&{ {x}}&{{ {x + a}}} \end{array}} \right|{ { = 0}}\quad \;\;\;\;{ {[}}{{ {R}}_1} \to {{ {R}}_1}{ { + }}{{ {R}}_2}{ { + }}{{ {R}}_3}{ {]}}\]
\[{ { = (3x + a)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {1}}&{ {1}} \\ { {x}}&{{ {x + 2}}}&{ {x}} \\ { {x}}&{ {x}}&{{ {x + a}}} \end{array}} \right|{ { = 0}}\] [\[{{ {R}}_1}\] से सामान्य रूप मे \[{ {(3x + a)}}\] ले]
\[\begin{align} { { = (3x + a)}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {0}} \\ { {x}}&{ {a}}&{ {0}} \\ { {x}}&{ {0}}&{ {a}} \end{array}} \right|{ { = 0 }}\quad { {[}}{{ {C}}_2} \to {{ {C}}_2}{ { - }}{{ {C}}_1}{ {, }}{{ {C}}_3} \to {{ {C}}_3}{ { - }}{{ {C}}_1}{ {]}} \hfill \\ { { = (3x + a)}}\left[ {{{ {a}}^{ {2}}}{ { - 0}}} \right] \hfill \\ { { = 0}} \hfill \\ \end{align} \]
\[\begin{align} { { = }}{{ {a}}^{ {2}}}{ {(3x + a) = 0}} \hfill \\ \Rightarrow { { (3x + a) = 0 }}\quad { {[}}\because { { a}} \ne { {0]}} \hfill \\ \Rightarrow { { x = - }}\dfrac{{ {a}}}{{ {3}}} \hfill \\ \end{align} \]
6. सिद्ध कीजिए की \[\left| {\begin{array}{*{20}{c}} {{{ {a}}^{ {2}}}}&{{ {bc}}}&{{ {ac + }}{{ {c}}^{ {2}}}} \\ {{{ {a}}^{ {2}}}{ { + ab}}}&{{{ {b}}^{ {2}}}}&{{ {ac}}} \\ {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + bc}}}&{{{ {c}}^{ {2}}}} \end{array}} \right|{ { = 4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}}\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} {{{ {a}}^{ {2}}}}&{{ {bc}}}&{{ {ac + }}{{ {c}}^{ {2}}}} \\ {{{ {a}}^{ {2}}}{ { + ab}}}&{{{ {b}}^{ {2}}}}&{{ {ac}}} \\ {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + bc}}}&{{{ {c}}^{ {2}}}} \end{array}} \right|{ { = }}\left| {\begin{array}{*{20}{c}} {{ {2}}{{ {a}}^{ {2}}}}&{ {0}}&{{ {2ac}}} \\ {{{ {a}}^{ {2}}}{ { + ab}}}&{{{ {b}}^{ {2}}}}&{{ {ac}}} \\ {{ {ab}}}&{{{ {b}}^{ {2}}}{ { + bc}}}&{{{ {c}}^{ {2}}}} \end{array}} \right|\]
\[{ { = abc}}\left| {\begin{array}{*{20}{c}} {{ {2a}}}&{ {0}}&{{ {2a}}} \\ {{ {a + b}}}&{ {b}}&{ {a}} \\ { {b}}&{{ {b + c}}}&{ {c}} \end{array}} \right|{ { [}}{{ {R}}_1} \to {{ {R}}_2}{ { + }}{{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]
\[\begin{align} { { = abc}}\left| {\begin{array}{*{20}{c}} {{ {2a}}}&{ {0}}&{ {0}} \\ {{ {a + b}}}&{ {b}}&{{ { - b}}} \\ { {b}}&{{ {b + c}}}&{{ {c - b}}} \end{array}} \right| \hfill \\ { { = (abc)2a[b(c - a) - b(b + c)]}} \hfill \\ { { = 2}}{{ {a}}^{ {2}}}{ {bc}}\left[ {{ {bc - }}{{ {b}}^{ {2}}}{ { + }}{{ {b}}^{ {2}}}{ { + bc}}} \right] \hfill \\ { { = 2}}{{ {a}}^{ {2}}}{ {bc[2bc]}} \hfill \\ { { = 4}}{{ {a}}^{ {2}}}{{ {b}}^{ {2}}}{{ {c}}^{ {2}}} \hfill \\ \end{align} \]
7. \[{{ {A}}^{{ { - 1}}}}{ { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}}&{ {1}} \\ {{ { - 15}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 2}}}&{ {2}} \end{array}} \right]\] और \[{ {B = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {2}}&{{ { - 2}}} \\ {{ { - 1}}}&{ {3}}&{ {0}} \\ { {0}}&{{ { - 2}}}&{ {1}} \end{array}} \right]\] हो तो \[{{ {(AB)}}^{{ { - 1}}}}\] का मान ज्ञात कीजिए।
उत्तर: \[{ {B = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{ {2}}&{{ { - 2}}} \\ {{ { - 1}}}&{ {3}}&{ {0}} \\ { {0}}&{{ { - 2}}}&{ {1}} \end{array}} \right]\]
\[\begin{array}{*{20}{c}} {{{ {B}}_{11}}{ { = 3}}}&{{{ {B}}_{12}}{ { = 1}}}&{{{ {B}}_{13}}{ { = 2}}} \\ {{{ {B}}_{21}}{ { = 2}}}&{{{ {B}}_{22}}{ { = 1}}}&{{{ {B}}_{23}}{ { = 2}}} \\ {{{ {B}}_{31}}{ { = 6}}}&{{{ {B}}_{32}}{ { = 2}}}&{{{ {B}}_{33}}{ { = 5}}} \end{array}\]
\[\begin{align} {{ {B}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|B|}}}}{ { adj B}} \hfill \\ { { = }}\dfrac{{ {1}}}{{ {1}}}\left[ {\begin{array}{*{20}{l}} {{{ {B}}_{{ {11}}}}}&{{{ {B}}_{{ {12}}}}}&{{{ {B}}_{{ {13}}}}} \\ {{{ {B}}_{{ {21}}}}}&{{{ {B}}_{{ {22}}}}}&{{{ {B}}_{{ {23}}}}} \\ {{{ {B}}_{{ {31}}}}}&{{{ {B}}_{{ {32}}}}}&{{{ {B}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{ {2}}&{ {6}} \\ { {1}}&{ {1}}&{ {2}} \\ { {2}}&{ {2}}&{ {5}} \end{array}} \right] \hfill \\ \therefore \;{{ {(AB)}}^{{ { - 1}}}}{ { = }}{{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}} \hfill \\ {{ {(AB)}}^{{ { - 1}}}}{ { = }}{{ {B}}^{{ { - 1}}}}{{ {A}}^{{ { - 1}}}}{ { = }}\left[ {\begin{array}{*{20}{l}} { {3}}&{ {2}}&{ {6}} \\ { {1}}&{ {1}}&{ {2}} \\ { {2}}&{ {2}}&{ {5}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {3}}&{{ { - 1}}}&{ {1}} \\ {{ { - 15}}}&{ {6}}&{{ { - 5}}} \\ { {5}}&{{ { - 2}}}&{ {2}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} {{ {9 - 30 + 30}}}&{{ { - 3 + 12 - 12}}}&{{ {3 - 10 + 12}}} \\ {{ {3 - 15 + 10}}}&{{ { - 1 + 6 - 4}}}&{{ {1 - 5 + 4}}} \\ {{ {6 - 30 + 25}}}&{{ { - 2 + 12 - 10}}}&{{ {2 - 10 + 10}}} \end{array}} \right] \hfill \\ { { = }}\left[ {\begin{array}{*{20}{c}} { {9}}&{{ { - 3}}}&{ {5}} \\ {{ { - 2}}}&{ {1}}&{ {0}} \\ { {1}}&{ {0}}&{ {2}} \end{array}} \right] \hfill \\ \end{align} \]
8. मान लीजिए \[\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 2}}}&{ {1}} \\ {{ { - 2}}}&{ {3}}&{ {1}} \\ { {1}}&{ {1}}&{ {5}} \end{array}} \right]\] हो तो सत्यापित कीजिए की
(i) \[{{ {[adjA]}}^{{ { - 1}}}}{ { = adj}}\left( {{{ {A}}^{{ { - 1}}}}} \right)\]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 2}}}&{ {1}} \\ {{ { - 2}}}&{ {3}}&{ {1}} \\ { {1}}&{ {1}}&{ {5}} \end{array}} \right]\]
\[\left| { {A}} \right|{ { = 1(15 - 1) + 2( - 10 - 1) + 1( - 2 - 3) = - 13 }} \ne { { 0 }} \Rightarrow { { }}{{ {A}}^{{ { - 1}}}}\]
\[\begin{array}{*{20}{c}} {{{ {A}}_{11}}{ { = 14}}}&{{{ {A}}_{12}}{ { = 11}}}&{{{ {A}}_{13}}{ { = - 5}}} \\ {{{ {A}}_{21}}{ { = 11}}}&{{{ {A}}_{22}}{ { = 4}}}&{{{ {A}}_{23}}{ { = - 3}}} \\ {{{ {A}}_{31}}{ { = - 5}}}&{{{ {A}}_{32}}{ { = - 3}}}&{{{ {A}}_{33}}{ { = - 1}}} \end{array}\]
\[{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ { adj A = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {13}}}}} \\ {{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {23}}}}} \\ {{{ {A}}_{{ {31}}}}}&{{{ {A}}_{{ {32}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}} {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\ {{ {11}}}&{ {4}}&{{ { - 3}}} \\ {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right]\] ………………..(1)
\[\begin{align} { {B = adjA}} \hfill \\ { {B = }}\left[ {\begin{array}{*{20}{c}} {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\ {{ {11}}}&{ {4}}&{{ { - 3}}} \\ {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right] \hfill \\ \left| { {B}} \right|{ { = 14( - 4 - 9) - 11( - 11 - 15) - 5( - 33 + 20) = - 182 + 286 + 65 = 169 }} \ne { { 0}}\; \Rightarrow \;{{ {B}}^{{ { - 1}}}} \hfill \\ \end{align} \]
\[\begin{array}{*{20}{c}} {{{ {B}}_{{ {11}}}}{ { = - 13}}}&{{{ {B}}_{{ {12}}}}{ { = 26}}}&{{{ {B}}_{{ {13}}}}{ { = - 13}}} \\ {{{ {B}}_{{ {21}}}}{ { = 26}}}&{{{ {B}}_{{ {22}}}}{ { = - 39}}}&{{{ {B}}_{{ {23}}}}{ { = - 13}}} \\ {{{ {B}}_{{ {31}}}}{ { = - 13}}}&{{{ {B}}_{{ {32}}}}{ { = - 13}}}&{{{ {B}}_{{ {33}}}}{ { = - 65}}} \end{array}\]
\[{{ {B}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|B|}}}}{ {adjB = }}\dfrac{{ {1}}}{{{ {|B|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {B}}_{{ {11}}}}}&{{{ {B}}_{{ {12}}}}}&{{{ {B}}_{{ {13}}}}} \\ {{{ {B}}_{{ {21}}}}}&{{{ {B}}_{{ {22}}}}}&{{{ {B}}_{{ {23}}}}} \\ {{{ {B}}_{{ {31}}}}}&{{{ {B}}_{{ {32}}}}}&{{{ {B}}_{{ {33}}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {169}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 13}}}&{{ {26}}}&{{ { - 13}}} \\ {{ {26}}}&{{ { - 39}}}&{{ { - 13}}} \\ {{ { - 13}}}&{{ { - 13}}}&{{ { - 65}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {2}}&{{ { - 3}}}&{{ { - 1}}} \\ {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}} \end{array}} \right]\]
\[{ {adj}}{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {2}}&{{ { - 3}}}&{{ { - 1}}} \\ {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}} \end{array}} \right]\] …………………..(2)
\[\begin{align} { {C = }}{{ {A}}^{{ { - 1}}}} \hfill \\ { {C = }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}} {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\ {{ {11}}}&{ {4}}&{{ { - 3}}} \\ {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{array}{*{20}{c}} {{{ {C}}_{{ {11}}}}{ { = 3}}}&{{{ {C}}_{{ {12}}}}{ { = 1}}}&{{{ {C}}_{{ {13}}}}{ { = 2}}} \\ {{{ {C}}_{21}}{ { = 2}}}&{{{ {C}}_{22}}{ { = 1}}}&{{{ {C}}_{23}}{ { = 2}}} \\ {{{ {C}}_{31}}{ { = 6}}}&{{{ {C}}_{32}}{ { = 2}}}&{{{ {C}}_{33}}{ { = 5}}} \end{array}\]
\[{ {adjC = }}\dfrac{{ {1}}}{{ {1}}}\left[ {\begin{array}{*{20}{l}} {{{ {C}}_{{ {11}}}}}&{{{ {C}}_{{ {12}}}}}&{{{ {C}}_{{ {13}}}}} \\ {{{ {C}}_{{ {21}}}}}&{{{ {C}}_{{ {22}}}}}&{{{ {C}}_{{ {23}}}}} \\ {{{ {C}}_{{ {31}}}}}&{{{ {C}}_{{ {32}}}}}&{{{ {C}}_{{ {33}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\ {\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\ {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {5}}}{{{ {13}}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {2}}&{{ { - 3}}}&{{ { - 1}}} \\ {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}} \end{array}} \right]\]
\[{ {adjC = adj}}\left( {{{ {A}}^{{ { - 1}}}}} \right){ { = }}\dfrac{{ {1}}}{{{ {13}}}}\left[ {\begin{array}{*{20}{c}} {{ { - 1}}}&{ {2}}&{{ { - 1}}} \\ { {2}}&{{ { - 3}}}&{{ { - 1}}} \\ {{ { - 1}}}&{{ { - 1}}}&{{ { - 5}}} \end{array}} \right]\] ………………..(3)
समीकरण (2) और (3) से हमारे पास है \[{{ {(adjA)}}^{{ { - 1}}}}{ { = adj}}\left( {{{ {A}}^{{ { - 1}}}}} \right)\]
(ii) \[{\left( {{{ {A}}^{{ { - 1}}}}} \right)^{{ { - 1}}}}{ { = A}}\]
उत्तर: समीकरण (1) से हमारे पास
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}} {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\ {{ {11}}}&{ {4}}&{{ { - 3}}} \\ {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right] \hfill \\ { {D = }}{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ { - 13}}}}\left[ {\begin{array}{*{20}{c}} {{ {14}}}&{{ {11}}}&{{ { - 5}}} \\ {{ {11}}}&{ {4}}&{{ { - 3}}} \\ {{ { - 5}}}&{{ { - 3}}}&{{ { - 1}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {\dfrac{{{ {14}}}}{{{ {13}}}}}&{\dfrac{{{ {11}}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {5}}}{{{ {13}}}}} \\ {\dfrac{{{ {11}}}}{{{ {13}}}}}&{\dfrac{{ {4}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}} \\ {{ { - }}\dfrac{{ {5}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \end{array}} \right] \hfill \\ \left| { {D}} \right|{ { = - }}{\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)^{ {3}}}{ {[14( - 4 - 9) - 11( - 11 - 15) - 5( - 33 + 20)]}} \hfill \\ { { = - }}{\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)^{ {3}}}{ {( - 182 + 286 + 65) = - }}{\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)^{ {3}}}{ {169 = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right){ { }} \ne { { 0 }} \Rightarrow { { }}{{ {D}}^{{ { - 1}}}} \hfill \\ \end{align} \]
\[\begin{array}{*{20}{l}} {{{ {D}}_{11}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{12}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{13}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)} \\ {{{ {D}}_{21}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{22}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{23}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)} \\ {{{ {D}}_{31}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{23}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)}&{{{ {D}}_{33}}{ { = - }}\left( {\dfrac{{ {1}}}{{{ {13}}}}} \right)} \end{array}\]
\[\begin{align} {{ {D}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|D|}}}}{ {adjD = }}\dfrac{{ {1}}}{{{ {|D|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {D}}_{{ {11}}}}}&{{{ {D}}_{{ {12}}}}}&{{{ {D}}_{{ {13}}}}} \\ {{{ {D}}_{{ {21}}}}}&{{{ {D}}_{{ {22}}}}}&{{{ {D}}_{{ {23}}}}} \\ {{{ {D}}_{{ {31}}}}}&{{{ {D}}_{{ {32}}}}}&{{{ {D}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ { - 1}}}}\left[ {\begin{array}{*{20}{c}} {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\ {\dfrac{{ {2}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {3}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}} \\ {{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {1}}}{{{ {13}}}}}&{{ { - }}\dfrac{{ {5}}}{{{ {13}}}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ { - 2}}}&{ {1}} \\ {{ { - 2}}}&{ {3}}&{ {1}} \\ { {1}}&{ {1}}&{ {5}} \end{array}} \right] \hfill \\ {{ {D}}^{{ { - 1}}}}{ { = }}{\left( {{{ {A}}^{{ { - 1}}}}} \right)^{{ { - 1}}}}{ { = }}\left| {\begin{array}{*{20}{c}} { {1}}&{{ { - 2}}}&{ {1}} \\ {{ { - 2}}}&{ {3}}&{ {1}} \\ { {1}}&{ {1}}&{ {5}} \end{array}} \right|{ { = A}} \hfill \\ \end{align} \]
9. \[\left| {\begin{array}{*{20}{c}} { {x}}&{ {y}}&{{ {x + y}}} \\ { {y}}&{{ {x + y}}}&{ {x}} \\ {{ {x + y}}}&{ {x}}&{ {y}} \end{array}} \right|\] का मान ज्ञात कीजिए।
उत्तर: \[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {2(x + y)}}}&{ {y}}&{{ {x + y}}} \\ {{ {2(x + y)}}}&{{ {x + y}}}&{ {x}} \\ {{ {2(x + y)}}}&{ {x}}&{ {y}} \end{array}} \right|{ { [}}{{ {C}}_1} \to {{ {C}}_1}{ { + }}{{ {C}}_2}{ { + }}{{ {C}}_3}{ {]}}\]
\[{ { = 2(x + y)}}\left| {\begin{array}{*{20}{c}} { {1}}&{ {y}}&{{ {x + y}}} \\ { {1}}&{{ {x + y}}}&{ {x}} \\ { {1}}&{ {x}}&{ {y}} \end{array}} \right|\] [\[{{ {C}}_{ {1}}}\] से सामान्य के रूप मे \[{ {2(x + y)}}\] लेना]
\[\begin{align} { { = 2(x + y)}}\left| {\begin{array}{*{20}{c}} { {0}}&{{ { - x}}}&{ {y}} \\ { {0}}&{ {y}}&{{ {x - y}}} \\ { {1}}&{ {x}}&{ {y}} \end{array}} \right|\;\;\;\;\;\;\,\;\;\;\;\;\;\left[ {{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}} \right.{ {]}} \hfill \\ { { = 2(x + y)[( - x)(x - y) - y \times y]}} \hfill \\ { { = 2(x + y)}}\left( {{ { - }}{{ {x}}^{ {2}}}{ { + xy - }}{{ {y}}^{ {2}}}} \right) \hfill \\ { { = - 2(x + y)}}\left( {{{ {x}}^{ {2}}}{ { - xy + }}{{ {y}}^{ {2}}}} \right) \hfill \\ { { = - 2}}\left( {{{ {x}}^{ {2}}}{ { + }}{{ {y}}^{ {2}}}} \right) \hfill \\ \end{align} \]
10. \[\left| {\begin{array}{*{20}{c}} { {1}}&{ {x}}&{{ {x + y}}} \\ { {1}}&{{ {x + y}}}&{ {x}} \\ { {1}}&{ {x}}&{ {y}} \end{array}} \right|\] का मान ज्ञात कीजिए।
उत्तर: \[{ { = }}\left| {\begin{array}{*{20}{c}} { {0}}&{{ { - y}}}&{ {0}} \\ { {0}}&{ {y}}&{{ { - x}}} \\ { {1}}&{ {x}}&{{ {x + y}}} \end{array}} \right|{ { [}}{{ {R}}_1} \to {{ {R}}_1}{ { - }}{{ {R}}_2}{ {, }}{{ {R}}_2} \to {{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]
\[\begin{align} { { = [( - y)( - x) - y \times 0]}} \hfill \\ { { = xy}} \hfill \\ \end{align} \]
सारणिकों के गुणधर्मों का प्रयोग करके निम्नलिखित 11 से 15 तक प्रश्नों को सिद्ध कीजिए:
11. \[\left| {\begin{array}{*{20}{l}} { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{{ {\beta + \gamma }}} \\ { {\beta }}&{{{ {\beta }}^{ {2}}}}&{{ {\gamma + \alpha }}} \\ { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{{ {\alpha + \beta }}} \end{array}} \right|{ { = (\beta - \gamma )(\gamma - \alpha )(\alpha - \beta )(\alpha + \beta + \gamma )}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}} { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{{ {\beta + \gamma }}} \\ { {\beta }}&{{{ {\beta }}^{ {2}}}}&{{ {\gamma + \alpha }}} \\ { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{{ {\alpha + \beta }}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{l}} { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{{ {\alpha + \beta + \gamma }}} \\ { {\beta }}&{{{ {\beta }}^{ {2}}}}&{{ {\alpha + \beta + \gamma }}} \\ { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{{ {\alpha + \beta + \gamma }}} \end{array}} \right|{ { [}}{{ {C}}_3} \to {{ {C}}_3}{ { + }}{{ {C}}_2}{ {]}}\]
\[{ { = (\alpha + \beta + \gamma )}}\left| {\begin{array}{*{20}{l}} { {\alpha }}&{{{ {\alpha }}^{ {2}}}}&{ {1}} \\ { {\beta }}&{{{ {\beta }}^{ {2}}}}&{ {1}} \\ { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{ {1}} \end{array}} \right|\] [\[{{ {C}}_3}\] से सामान्य के रूप मे \[{ {(\alpha + \beta + \gamma )}}\] लेना]
\[{ { = (\alpha + \beta + \gamma )}}\left| {\begin{array}{*{20}{c}} {{ {\alpha - \beta }}}&{{{ {\alpha }}^{ {2}}}{ { - }}{{ {\beta }}^{ {2}}}}&{ {0}} \\ {{ {\beta - \gamma }}}&{{{ {\beta }}^{ {2}}}{ { - }}{{ {\gamma }}^{ {2}}}}&{ {0}} \\ { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{ {1}} \end{array}} \right|\quad \;\;\;\;\;\;\;\,\;{ {[}}{{ {R}}_1} \to {{ {R}}_1}{ { - }}{{ {R}}_2}{ {, }}{{ {R}}_2} \to {{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]
\[{ { = (\alpha + \beta + \gamma )(\alpha - \beta )(\beta - \gamma )}}\left| {\begin{array}{*{20}{c}} { {1}}&{{ {\alpha + \beta }}}&{ {0}} \\ { {1}}&{{ {\beta + \gamma }}}&{ {0}} \\ { {\gamma }}&{{{ {\gamma }}^{ {2}}}}&{ {1}} \end{array}} \right|\] [\[{{ {R}}_{ {1}}}\] से सामान्य के रूप मे \[{ {(\alpha - \beta )}}\] लेना और \[{{ {R}}_{ {2}}}\] से सामान्य रूप से \[{ {(\beta - \gamma )}}\] लेना]
\[\begin{align} { { = (\alpha + \beta + \gamma )(\alpha - \beta )(\beta - \gamma )[(\beta + \gamma ) - (\alpha + \beta )]}} \hfill \\ { { = (\alpha + \beta + \gamma )(\alpha - \beta )(\beta - \gamma )(\gamma - \alpha )}} \hfill \\ \end{align} \]
= RHS
12. \[\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{{ {1 + p}}{{ {x}}^{ {2}}}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{{ {1 + p}}{{ {y}}^{ {2}}}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{{ {1 + p}}{{ {z}}^{ {2}}}} \end{array}} \right|{ { = (1 + pxyz)(x - y)(y - z)(z - x)}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{{ {1 + p}}{{ {x}}^{ {2}}}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{{ {1 + p}}{{ {y}}^{ {2}}}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{{ {1 + p}}{{ {z}}^{ {2}}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{ {1}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{ {1}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{ {1}} \end{array}} \right|{ { + }}\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{{ {p}}{{ {x}}^{ {2}}}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{{ {p}}{{ {y}}^{ {2}}}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{{ {p}}{{ {z}}^{ {2}}}} \end{array}} \right|\]
\[{ { = ( - 1}}{{ {)}}^{ {1}}}\left| {\begin{array}{*{20}{l}} { {1}}&{{{ {x}}^{ {2}}}}&{ {x}} \\ { {1}}&{{{ {y}}^{ {2}}}}&{ {y}} \\ { {1}}&{{{ {z}}^{ {2}}}}&{ {z}} \end{array}} \right|{ { + p}}\left| {\begin{array}{*{20}{l}} { {x}}&{{{ {x}}^{ {2}}}}&{{{ {x}}^{ {3}}}} \\ { {y}}&{{{ {y}}^{ {2}}}}&{{{ {y}}^{ {3}}}} \\ { {z}}&{{{ {z}}^{ {2}}}}&{{{ {z}}^{ {3}}}} \end{array}} \right|\]
[\[{{ {C}}_3}\] से सामान्य के रूप मे \[{ {p}}\] लेना]
\[{ { = ( - 1}}{{ {)}}^{ {2}}}\left| {\begin{array}{*{20}{l}} { {1}}&{{{ {x}}^{ {2}}}}&{ {x}} \\ { {1}}&{{{ {y}}^{ {2}}}}&{ {y}} \\ { {1}}&{{{ {z}}^{ {2}}}}&{ {z}} \end{array}} \right|{ { + pxyz}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\ { {1}}&{ {y}}&{{{ {y}}^{ {2}}}} \\ { {1}}&{ {z}}&{{{ {z}}^{ {2}}}} \end{array}} \right|\] [\[{{ {R}}_1}{ {,}}{{ {R}}_2}{ {,}}{{ {R}}_3}\] से सामान्य के रूप मे \[{ {x,y,z}}\] लेना]
\[{ { = (1 + pxyz)}}\left| {\begin{array}{*{20}{l}} { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\ { {1}}&{ {y}}&{{{ {y}}^{ {2}}}} \\ { {1}}&{ {z}}&{{{ {z}}^{ {2}}}} \end{array}} \right|\]
[\[\,\left| {\begin{array}{*{20}{l}} { {1}}&{ {x}}&{{{ {x}}^{ {2}}}} \\ { {1}}&{ {y}}&{{{ {y}}^{ {2}}}} \\ { {1}}&{ {z}}&{{{ {z}}^{ {2}}}} \end{array}} \right|\] सामान्य के रूप मे लेना]
\[{ { = (1 + pxyz)}}\left| {\begin{array}{*{20}{c}} { {0}}&{{ {x - y}}}&{{{ {x}}^{ {2}}}{ { - }}{{ {y}}^{ {2}}}} \\ { {0}}&{{ {y - z}}}&{{{ {y}}^{ {2}}}{ { - }}{{ {z}}^{ {2}}}} \\ { {1}}&{ {z}}&{{{ {z}}^{ {2}}}} \end{array}} \right|\;\;\;\,\,\,{ { }}\,\;{ {[}}{{ {R}}_{ {1}}} \to {{ {R}}_{ {1}}}{ { - }}{{ {R}}_{ {2}}}{ {, }}{{ {R}}_{ {2}}} \to {{ {R}}_{ {2}}}{ { - }}{{ {R}}_{ {3}}}]\]
\[{ { = (1 + pxyz)(x - y)(y - z)}}\left| {\begin{array}{*{20}{c}} { {0}}&{ {1}}&{{ {x + y}}} \\ { {0}}&{ {1}}&{{ {y + z}}} \\ { {1}}&{ {z}}&{{{ {z}}^{ {2}}}} \end{array}} \right|\] [\[{{ {R}}_1}\] से सामान्य के रूप मे \[{ {(x - y)}}\] लेना और \[{{ {R}}_2}\] से सामान्य के रूप मे \[{ {y - z}}\] लेना]
\[\begin{align} { { = (1 + pxyz)(x - y)(y - z)[(y + z) - (x + y)]}} \hfill \\ { { = (1 + pxyz)(x - y)(y - z)(z - x)}} \hfill \\ \end{align} \]
= RHS
13. \[\left| {\begin{array}{*{20}{c}} {{ {3a}}}&{{ { - a + b}}}&{{ { - a + c}}} \\ {{ { - b + a}}}&{{ {3b}}}&{{ { - b + c}}} \\ {{ { - c + a}}}&{{ { - c + b}}}&{{ {3c}}} \end{array}} \right|{ { = 3(a + b + c)(ab + bc + ca)}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {3a}}}&{{ { - a + b}}}&{{ { - a + c}}} \\ {{ { - b + a}}}&{{ {3b}}}&{{ { - b + c}}} \\ {{ { - c + a}}}&{{ { - c + b}}}&{{ {3c}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {a + b + c}}}&{{ { - a + b}}}&{{ { - a + c}}} \\ {{ {a + b + c}}}&{{ {3b}}}&{{ { - b + c}}} \\ {{ {a + b + c}}}&{{ { - c + b}}}&{{ {3c}}} \end{array}} \right|{ { [}}{{ {C}}_1} \to {{ {C}}_1}{ { + }}{{ {C}}_2}{ { + }}{{ {C}}_3}{ {]}}\]
\[{ { = (a + b + c)}}\left| {\begin{array}{*{20}{c}} { {1}}&{{ { - a + b}}}&{{ { - a + c}}} \\ { {1}}&{{ {3b}}}&{{ { - b + c}}} \\ { {1}}&{{ { - c + b}}}&{{ {3c}}} \end{array}} \right|\] [\[{{ {C}}_{ {1}}}\] से सामान्य के रूप मे \[{ {(a + b + c)}}\] लेना]
\[{ { = (a + b + c)}}\left| {\begin{array}{*{20}{c}} { {0}}&{{ { - a - 2b}}}&{{ { - a + c}}} \\ { {0}}&{{ {2b + c}}}&{{ { - b - 2c}}} \\ { {1}}&{{ { - c + b}}}&{{ {3c}}} \end{array}} \right|{ { [}}{{ {R}}_1} \to {{ {R}}_1}{ { - }}{{ {R}}_2}{ {, }}{{ {R}}_2} \to {{ {R}}_2}{ { - }}{{ {R}}_3}{ {]}}\]
\[\begin{align} { { = (a + b + c)[( - a - 2b)( - b - 2c) - (2b + c)( - a + b)]}} \hfill \\ { { = (a + b + c)}}\left( {{ {ab + 2ac + 2}}{{ {b}}^{ {2}}}{ { + 4bc}}} \right.\left. {{ { - }}\left( {{ { - 2ab + 2}}{{ {b}}^{ {2}}}{ { = - ac + bc}}} \right)} \right) \hfill \\ { { = (a + b + c)(3ab + 3bc + 3ca)}} \hfill \\ { { = 3(a + b + c)(ab + bc + ca)}} \hfill \\ \end{align} \]
= RHS
14. \[\left| {\begin{array}{*{20}{c}} { {1}}&{{ {1 + p}}}&{{ {1 + p + q}}} \\ { {2}}&{{ {3 + 2p}}}&{{ {4 + 3p + 2q}}} \\ { {3}}&{{ {6 + 3p}}}&{{ {10 + 6p + 3q}}} \end{array}} \right|{ { = 1}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} { {1}}&{{ {1 + p}}}&{{ {1 + p + q}}} \\ { {2}}&{{ {3 + 2p}}}&{{ {4 + 3p + 2q}}} \\ { {3}}&{{ {6 + 3p}}}&{{ {10 + 6p + 3q}}} \end{array}} \right|\]
\[\begin{align} { { = }}\left| {\begin{array}{*{20}{c}} { {1}}&{{ {1 + p}}}&{{ {1 + p + q}}} \\ { {0}}&{ {1}}&{{ {2 + p}}} \\ { {0}}&{ {3}}&{{ {7 + 3p}}} \end{array}} \right|{ { [}}{{ {R}}_2} \to {{ {R}}_2}{ { - 2}}{{ {R}}_1}{ {, }}{{ {R}}_3} \to {{ {R}}_3}{ { - 3}}{{ {R}}_1}{ {]}} \hfill \\ { { = 1(1(7 + 3p) - (3)(2 + p))}} \hfill \\ { { = 7 + 3p - 6 - 3p}} \hfill \\ { { = 1}} \hfill \\ \end{align} \]
= RHS
15. \[\left| {\begin{array}{*{20}{c}} {{ {sin\alpha }}}&{{ {cos\alpha }}}&{{ {cos(\alpha + \delta )}}} \\ {{ {sin\beta }}}&{{ {cos\beta }}}&{{ {cos(\beta + \delta )}}} \\ {{ {sin\gamma }}}&{{ {cos\gamma }}}&{{ {cos(\gamma + \delta )}}} \end{array}} \right|{ { = 0}}\]
उत्तर: LHS = \[\left| {\begin{array}{*{20}{c}} {{ {sin\alpha }}}&{{ {cos\alpha }}}&{{ {cos(\alpha + \delta )}}} \\ {{ {sin\beta }}}&{{ {cos\beta }}}&{{ {cos(\beta + \delta )}}} \\ {{ {sin\gamma }}}&{{ {cos\gamma }}}&{{ {cos(\gamma + \delta )}}} \end{array}} \right|\]
\[\begin{align} { { = }}\left| {\begin{array}{*{20}{l}} {{ {sin\alpha }}}&{{ {cos\alpha cos\delta - sin\alpha sin\delta }}}&{{ {cos(\alpha + \delta )}}} \\ {{ {sin\beta }}}&{{ {cos\beta cos\delta - sin\beta sin\delta }}}&{{ {cos(\beta + \delta )}}} \\ {{ {sin\gamma }}}&{{ {cos\gamma cos\delta - sin\gamma sin\delta }}}&{{ {cos(\gamma + \delta )}}} \end{array}} \right|{ { [}}{{ {C}}_2} \to { {cos\delta }}{{ {C}}_2}{ { - sin\delta }}{{ {C}}_1}{ {]}} \hfill \\ { { = }}\left| {\begin{array}{*{20}{l}} {{ {sin\alpha }}}&{{ {cos(\alpha + \delta )}}}&{{ {cos(\alpha + \delta )}}} \\ {{ {sin\beta }}}&{{ {cos(\beta + \delta )}}}&{{ {cos(\beta + \delta )}}} \\ {{ {sin\gamma }}}&{{ {cos(\gamma + \delta )}}}&{{ {cos(\gamma + \delta )}}} \end{array}} \right|\quad \;\;\,\;\;\,\,\,\left[ {\because \;{{ {C}}_{ {2}}}{ { = }}{{ {C}}_{ {3}}}} \right] \hfill \\ { { = 0}} \hfill \\ \end{align} \]
= RHS
16. निम्नलिखित समीकरण निकाय को हल कीजिए:
\[\dfrac{{ {2}}}{{ {x}}}{ { + }}\dfrac{{ {3}}}{{ {y}}}{ { + }}\dfrac{{{ {10}}}}{{ {z}}}{ { = 4, }}\dfrac{{ {4}}}{{ {x}}}{ { - }}\dfrac{{ {6}}}{{ {y}}}{ { + }}\dfrac{{ {5}}}{{ {z}}}{ { = 1, }}\dfrac{{ {6}}}{{ {x}}}{ { + }}\dfrac{{ {9}}}{{ {y}}}{ { - }}\dfrac{{{ {20}}}}{{ {z}}}{ { = 2}}\]
उत्तर: समीकरणों की इस प्रणाली को \[{ {AX}}\,{ { = }}\;{ {B}}\] के रूप मे लिखा जा सकता है:
\[\begin{align} { {A = }}\left[ {\begin{array}{*{20}{c}} { {2}}&{ {3}}&{{ {10}}} \\ { {4}}&{{ { - 6}}}&{ {5}} \\ { {6}}&{ {9}}&{{ { - 7}}} \end{array}} \right]{ {, X = }}\left[ {\begin{array}{*{20}{l}} {\dfrac{{ {1}}}{{ {x}}}} \\ {\dfrac{{ {1}}}{{ {y}}}} \\ {\dfrac{{ {1}}}{{ {z}}}} \end{array}} \right]{ {, B = }}\left[ {\begin{array}{*{20}{l}} { {4}} \\ { {1}} \\ { {2}} \end{array}} \right] \hfill \\ { {|A| = 2(120 - 45) - 3( - 80 - 30) + 10(36 + 36)}} \hfill \\ { { = 150 + 330 + 720 = 1200 }} \ne { { 0 }} \Rightarrow { { }}{{ {A}}^{{ { - 1}}}} \hfill \\ \end{align} \]
\[\begin{array}{*{20}{c}} {{{ {A}}_{11}}{ { = 75}}}&{{{ {A}}_{12}}{ { = 110}}}&{{{ {A}}_{13}}{ { = 72}}} \\ {{{ {A}}_{21}}{ { = 150}}}&{{{ {A}}_{22}}{ { = - 100}}}&{{{ {A}}_{23}}{ { = 0}}} \\ {{{ {A}}_{31}}{ { = 75}}}&{{{ {A}}_{32}}{ { = 30}}}&{{{ {A}}_{33}}{ { = - 24}}} \end{array}\]
\[{{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {adjA = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {13}}}}} \\ {{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {23}}}}} \\ {{{ {A}}_{{ {31}}}}}&{{{ {A}}_{{ {32}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{c}} {{ {75}}}&{{ {150}}}&{{ {75}}} \\ {{ {110}}}&{{ { - 100}}}&{{ {30}}} \\ {{ {72}}}&{ {0}}&{{ { - 24}}} \end{array}} \right]\]
\[\begin{align} { {X = }}{{ {A}}^{{ { - 1}}}}{ {B}} \hfill \\ \left[ {\begin{array}{*{20}{l}} {\dfrac{{ {1}}}{{ {x}}}} \\ {\dfrac{{ {1}}}{{ {y}}}} \\ {\dfrac{{ {1}}}{{ {z}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{c}} {{ {75}}}&{{ {150}}}&{{ {75}}} \\ {{ {110}}}&{{ { - 100}}}&{{ {30}}} \\ {{ {72}}}&{ {0}}&{{ { - 24}}} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} { {4}} \\ { {1}} \\ { {2}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} \left[ {\begin{array}{*{20}{l}} {\dfrac{{ {1}}}{{ {x}}}} \\ {\dfrac{{ {1}}}{{ {y}}}} \\ {\dfrac{{ {1}}}{{ {z}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{c}} {{ {300 + 150 + 150}}} \\ {{ {440 - 100 + 60}}} \\ {{ {288 + 0 - 48}}} \end{array}} \right] \hfill \\ \left[ {\begin{array}{*{20}{l}} {\dfrac{{ {1}}}{{ {x}}}} \\ {\dfrac{{ {1}}}{{ {y}}}} \\ {\dfrac{{ {1}}}{{ {z}}}} \end{array}} \right]{ { = }}\dfrac{{ {1}}}{{{ {1200}}}}\left[ {\begin{array}{*{20}{l}} {{ {600}}} \\ {{ {400}}} \\ {{ {240}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{c}} {\dfrac{{ {1}}}{{ {2}}}} \\ { {1}} \\ { {3}} \\ { {1}} \\ { {5}} \end{array}} \right] \hfill \\ \end{align} \]
\[\begin{align} \Rightarrow \;\dfrac{{ {1}}}{{ {x}}}{ { = }}\dfrac{{ {1}}}{{ {2}}}{ {, }}\dfrac{{ {1}}}{{ {y}}}{ { = }}\dfrac{{ {1}}}{{ {3}}}{ {, }}\dfrac{{ {1}}}{{ {z}}}{ { = }}\dfrac{{ {1}}}{{ {5}}} \hfill \\ \Rightarrow \;{ {x = 2, y = 3, z = 5}} \hfill \\ \end{align} \]
निम्नलिखित प्रश्नों 17 से 19 मे सही उत्तर का चुनाव कीजिए।
17. यदि \[{ {a, b, c}}\] समातर श्रहई मे हो तो सारणिक \[\left| {\begin{array}{*{20}{c}} {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\ {{ {x + 3}}}&{{ {x + 4}}}&{{ {x + 2b}}} \\ {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}} \end{array}} \right|\] का मान होगा:
(a) \[{ {0}}\]
(b) \[{ {1}}\]
(c) \[{ {x}}\]
(d) \[{ {2x}}\]
उत्तर: \[\left| {\begin{array}{*{20}{c}} {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\ {{ {x + 3}}}&{{ {x + 4}}}&{{ {x + 2b}}} \\ {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}} \end{array}} \right|\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\ { {0}}&{ {0}}&{{ {2(2b - a - c)}}} \\ {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}} \end{array}} \right|\,\,\,\,\;\;\,\quad \left[ {{{ {R}}_{ {2}}} \to { {2}}{{ {R}}_{ {2}}}{ { - }}} \right.\left( {{{ {R}}_{ {1}}}{ { + }}{{ {R}}_{ {3}}}} \right){ {]}}\]
\[{ { = }}\left| {\begin{array}{*{20}{c}} {{ {x + 2}}}&{{ {x + 3}}}&{{ {x + 2a}}} \\ { {0}}&{ {0}}&{ {0}} \\ {{ {x + 4}}}&{{ {x + 5}}}&{{ {x + 2c}}} \end{array}} \right|\] [\[{ {a,b,c}}\] समांतर श्रेणी मे है]
\[{ { = 0}}\]
अतः विकल्प (a) सही है।
18. यदि \[{ {x, y, z}}\] शुनएतर वास्तविक संखयाए हो तो आव्यूह \[{ {A = }}\left| {\begin{array}{*{20}{l}} { {x}}&{ {0}}&{ {0}} \\ { {0}}&{ {y}}&{ {0}} \\ { {0}}&{ {0}}&{ {z}} \end{array}} \right|\] का वयुक्तकर्म है:
(a) \[\left[ {\begin{array}{*{20}{c}} {{{ {x}}^{{ { - 1}}}}}&{ {0}}&{ {0}} \\ { {0}}&{{{ {y}}^{{ { - 1}}}}}&{ {0}} \\ { {0}}&{ {0}}&{{{ {z}}^{{ { - 1}}}}} \end{array}} \right]\]
(b) \[{ {xyz}}\left[ {\begin{array}{*{20}{c}} {{{ {x}}^{{ { - 1}}}}}&{ {0}}&{ {0}} \\ { {0}}&{{{ {y}}^{{ { - 1}}}}}&{ {0}} \\ { {0}}&{ {0}}&{{{ {z}}^{{ { - 1}}}}} \end{array}} \right]\]
(c) \[\dfrac{{ {1}}}{{{ {xyz}}}}\left[ {\begin{array}{*{20}{l}} { {x}}&{ {0}}&{ {0}} \\ { {0}}&{ {y}}&{ {0}} \\ { {0}}&{ {0}}&{ {z}} \end{array}} \right]\]
(d) \[\dfrac{{ {1}}}{{{ {xyz}}}}\left[ {\begin{array}{*{20}{l}} { {1}}&{ {0}}&{ {0}} \\ { {0}}&{ {1}}&{ {0}} \\ { {0}}&{ {0}}&{ {1}} \end{array}} \right]\]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{l}} { {x}}&{ {0}}&{ {0}} \\ { {0}}&{ {y}}&{ {0}} \\ { {0}}&{ {0}}&{ {z}} \end{array}} \right]\]
\[\begin{align} { {|A| = x(yz - 0) - 0(0 - 0) + 0(0 + 0)}} \hfill \\ { { = xyz }} \ne { { 0 }} \Rightarrow { { }}{{ {A}}^{{ { - 1}}}} \hfill \\ \end{align} \]
\[\begin{array}{*{20}{c}} {{{ {A}}_{{ {11}}}}{ { = yz }}}&{{{ {A}}_{{ {12}}}}{ { = 0 }}}&{{{ {A}}_{{ {13}}}}{ { = 0 }}} \\ {{{ {A}}_{{ {21}}}}{ { = 0 }}}&{{{ {A}}_{{ {22}}}}{ { = xz }}}&{{{ {A}}_{{ {23}}}}{ { = 0}}} \\ {{{ {A}}_{{ {31}}}}{ { = 0 }}}&{{{ {A}}_{{ {32}}}}{ { = 0 }}}&{{{ {A}}_{{ {33}}}}{ { = xy }}} \end{array}\]
\[\begin{align} {{ {A}}^{{ { - 1}}}}{ { = }}\dfrac{{ {1}}}{{{ {|A|}}}}{ {adjA = }}\dfrac{{ {1}}}{{{ {|A|}}}}\left[ {\begin{array}{*{20}{l}} {{{ {A}}_{{ {11}}}}}&{{{ {A}}_{{ {12}}}}}&{{{ {A}}_{{ {13}}}}} \\ {{{ {A}}_{{ {21}}}}}&{{{ {A}}_{{ {22}}}}}&{{{ {A}}_{{ {23}}}}} \\ {{{ {A}}_{{ {31}}}}}&{{{ {A}}_{{ {32}}}}}&{{{ {A}}_{{ {33}}}}} \end{array}} \right] \hfill \\ { { = }}\dfrac{{ {1}}}{{{ {xyZ}}}}\left[ {\begin{array}{*{20}{c}} {{ {yz}}}&{ {0}}&{ {0}} \\ { {0}}&{{ {xz}}}&{ {0}} \\ { {0}}&{ {0}}&{{ {xy}}} \end{array}} \right]{ { = }}\left[ {\begin{array}{*{20}{l}} {\dfrac{{ {1}}}{{ {x}}}}&{ {0}}&{ {0}} \\ { {0}}&{\dfrac{{ {1}}}{{ {y}}}}&{ {0}} \\ { {0}}&{ {0}}&{\dfrac{{ {1}}}{{ {z}}}} \end{array}} \right] \hfill \\ \end{align} \]
\[{ { = }}\left[ {\begin{array}{*{20}{c}} {{{ {x}}^{{ { - 1}}}}}&{ {0}}&{ {0}} \\ { {0}}&{{{ {y}}^{{ { - 1}}}}}&{ {0}} \\ { {0}}&{ {0}}&{{{ {z}}^{{ { - 1}}}}} \end{array}} \right]\]
अतः विकल्प (a) सही है।
19. यदि \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ {sin\theta }}}&{ {1}} \\ {{ { - sin\theta }}}&{ {1}}&{{ {sin\theta }}} \\ {{ { - 1}}}&{{ { - sin\theta }}}&{ {1}} \end{array}} \right]\] जहा \[{ {0 }} \leqslant { { \theta }} \leqslant { { 2\pi }}\] हो तो:
(a) \[{ {det(A) = 0}}\]
(b) \[{ {det(A)}} \in { {(2,}}\infty { {)}}\]
(c) \[{ {det(A)}} \in { {(2,4)}}\]
(d) \[{ {det(A)}} \in { {[2,4]}}\]
उत्तर: \[{ {A = }}\left[ {\begin{array}{*{20}{c}} { {1}}&{{ {sin\theta }}}&{ {1}} \\ {{ { - sin\theta }}}&{ {1}}&{{ {sin\theta }}} \\ {{ { - 1}}}&{{ { - sin\theta }}}&{ {1}} \end{array}} \right]\]
\[\begin{align} { { = 1}}\left( {{ {1 + si}}{{ {n}}^{ {2}}}{ {\theta }}} \right){ { + sin\theta ( - sin\theta + sin\theta ) + 1}}\left( {{ {si}}{{ {n}}^{ {2}}}{ {\theta + 1}}} \right) \hfill \\ { { = 2}}\left( {{ {1 + si}}{{ {n}}^{ {2}}}{ {\theta }}} \right) \hfill \\ \end{align} \]
अब यह देखते हुए की \[{ {0}} \leqslant { {\theta }} \leqslant { {2\pi }}\]
\[ \Rightarrow { { 0}} \leqslant { {sin\theta }} \leqslant { {1}}\]
\[\begin{align} \Rightarrow \;{ {0}} \leqslant { {si}}{{ {n}}^{ {2}}}{ {\theta }} \leqslant { {1}} \hfill \\ \Rightarrow \;{ {1}} \leqslant { {1 + si}}{{ {n}}^{ {2}}}{ {\theta }} \leqslant { {2}} \hfill \\ \Rightarrow \;{ {2}} \leqslant { {2}}\left( {{ {1 + si}}{{ {n}}^{ {2}}}{ {\theta }}} \right) \leqslant { {4}} \hfill \\ \Rightarrow \;{ {det(A)}} \in { {[2,4]}} \hfill \\ \end{align} \]
अतः विकल्प (d) सही है।
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