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NCERT Solutions

Class 12

Maths In Hindi

Chapter 4 Determinants

NCERT Solutions for Class 12 Maths In Hindi Chapter 4 Determinants

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. $| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} |$

उत्तर: $| \begin{matrix} 2 & 4 \\ - 5 & - 1 \end{matrix} |$ = $2 (- 1) - 4 (- 5) = - 2 + 20 = 18$

2. (i) $| \begin{matrix} c o s θ & - s i n θ \\ s i n θ & c o s θ \end{matrix} |$

उत्तर: $| \begin{matrix} c o s θ & - s i n θ \\ s i n θ & c o s θ \end{matrix} |$ = $(c o s θ \times c o s θ) - (s i n θ \times (- s i n θ))$

$= c o s^{2} θ + s i n^{2} θ = 1$

(ii) $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

उत्तर: $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} | = (x^{2} - x + 1) \times (x + 1) - (x + 1) \times (x - 1)$

$\begin{array}{r} = x^{3} + x^{2} - x^{2} - x + x + 1 - (x^{2} + x - x - 1) \\ = x^{3} - x^{2} + 2 \end{array}$

3. यदि $A = [\begin{array}{l} 1 & 2 \\ 4 & 2 \end{array}]$ तो दिखाइए $| 2 A | = 4 | A ∣$

उत्तर: $A = [\begin{array}{l} 1 & 2 \\ 4 & 2 \end{array}]$

$| 2 A | = | \begin{array}{l} 2 & 4 \\ 8 & 4 \end{array} | = 2 (4) - 8 (4) = 8 - 32 = - 24$ .......(i)

$4 | A | = 4 | \begin{array}{l} 1 & 2 \\ 4 & 2 \end{array} | = 4 [1 (2) - 2 (4)] = 4 (2 - 8) = - 24$ .......(ii)

अतः (i) और (ii) से $| 2 A | = 4 | A ∣$

4. यदि $A = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}]$ हो, तो दिखाइए $| 3 A | = 27 | A ∣$

उत्तर: $A = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array}]$

$| 3 A | = | \begin{matrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{matrix} | = 3 (36 - 0) - 0 (0 - 0) + 3 (0 - 0) = 108$ .......(i)

$27 | A | = 27 | \begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{array} | = 27 [1 (4 - 0) - 0 (0 - 0) - 1 (0 - 0)] = 27 (4) = 108$ .......(ii)

अतः (i) और (ii) से $| 3 A | = 27 | A ∣$

5. निम्नलिखित सारणिकों का मान ज्ञात कीजिए:

(i) $| \begin{matrix} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$

उत्तर: $| \begin{matrix} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{matrix} | = 3 (0 - 5) + 1 (0 + 3) + 2 (0 - 0)$

$\begin{array}{r} = - 15 + 3 - 0 \\ = - 12 \end{array}$

(ii) $| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} |$

उत्तर: $| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} | = 3 (1 + 6) + 4 (1 + 4) + 5 (3 - 2)$

$\begin{array}{r} = 21 + 20 + 5 \\ = 46 \end{array}$

(iii) $| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} |$

उत्तर: $| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} |$ = $0 (0 + 9) - 1 (0 - 6) + 2 (- 3 - 0)$

$= 0 + 6 - 6 = 0$

(iv) $| \begin{matrix} 2 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$

उत्तर: $| \begin{matrix} 2 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix} | = 2 (0 - 5) + 1 (0 + 3) - 2 (0 - 6)$

$\begin{array}{r} = - 10 + 3 + 12 \\ = 5 \end{array}$

6. यदि $A = [\begin{array}{l} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 5 & 4 & 9 \end{array}]$ हो, तो ज्ञात कीजिए $| A |$

उत्तर: $| A | = | \begin{array}{l} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 5 & 4 & 9 \end{array} |$

$\begin{array}{r} = 1 (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5) \\ = 3 + 3 - 6 \\ = 0 \end{array}$

7. $x$ के मान ज्ञात कीजिए यदि

(i) $| \begin{array}{l} 2 & 4 \\ 5 & 1 \end{array} | = | \begin{matrix} 2 x & 4 \\ 6 & x \end{matrix} |$

उत्तर: $2 (1) - 5 (4) = 2 x (x) - 6 (4)$

$\begin{array}{r} 2 - 20 = 2 x^{2} - 24 \\ - 18 = 2 x^{2} - 24 \\ 2 x^{2} - 24 + 18 = 0 \end{array}$

$2 x^{2} - 6 = 0$

$\begin{array}{r} x^{2} = \frac{6}{2} = 3 \\ x = \pm \sqrt{3} \end{array}$

(ii) $| \begin{array}{l} 2 & 3 \\ 4 & 5 \end{array} | = | \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} |$

उत्तर: $2 (5) - 4 (3) = x (5) - 3 (2 x)$

$\begin{array}{r} 10 - 12 = 5 x - 6 x \\ - 2 = - x \\ x = 2 \end{array}$

8. यदि $| \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = | \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} |$ हो तो $x$ बराबर है

(a) $6$

(b) $\pm 6$

(d) $0$

उत्तर: $| \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = | \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} |$

$\begin{array}{r} x^{2} - 36 = 36 - 36 \\ x^{2} = 36 \\ x = \pm 6 \end{array}$

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. $| \begin{array}{l} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{array} | = 0$

उत्तर: $| \begin{array}{l} x & a & x + a \\ y & b & y + b \\ z & c & z + c \end{array} |$ = $| \begin{array}{l} x + a & a & x + a \\ y + b & b & y + b \\ z + c & c & z + c \end{array} | [∵ C_{1} \to C_{1} + C_{2}]$

$= 0 [∵ C_{1} = C_{3}]$

2. $| \begin{array}{l} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{array} | = 0$

उत्तर: $= | \begin{array}{l} 0 & b - c & c - a \\ 0 & c - a & a - b \\ 0 & a - b & b - c \end{array} | [C_{1} \to C_{1} + C_{2} + C_{3}]$

$= 0 [C_{1} = 0]$

3. $| \begin{array}{l} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array} | = 0$

उत्तर: $| \begin{array}{l} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array} | = | \begin{array}{l} 2 & 7 & 63 \\ 3 & 8 & 72 \\ 5 & 9 & 81 \end{array} | [C_{3} \to C_{3} - C_{1}]$

$= 9 | \begin{array}{l} 2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9 \end{array} |$ $C_{3}$ से $9$ उभयनिष्ट लेने पर]

$= 0 [C_{2} = C_{3}]$

4. $| \begin{array}{l} 1 & b c & a (b + c) \\ 1 & c a & b (c + a) \\ 1 & a b & c (a + b) \end{array} | = 0$

उत्तर: $| \begin{array}{l} 1 & b c & a (b + c) \\ 1 & c a & b (c + a) \\ 1 & a b & c (a + b) \end{array} | = | \begin{array}{l} 1 & b c & a b + b c + c a \\ 1 & c a & a b + b c + c a \\ 1 & a b & a b + b c + c a \end{array} |$ $[C_{3} \to C_{3} + C_{2}]$

$= (a b + b c + c a) | \begin{array}{l} 1 & b c & 1 \\ 1 & c a & 1 \\ 1 & a b & 1 \end{array} |$

( $C_{3}$ से $(a b + b c + c a)$ उभयनिष्ठ लेने पर)

$= 0 [C_{1} = C_{3}]$

5. $| \begin{array}{l} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{array} | = 2 | \begin{array}{l} a & p & x \\ b & q & y \\ c & r & z \end{array} |$

उत्तर: $| \begin{array}{l} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{array} | = 2 | \begin{array}{l} a & p & x \\ b & q & y \\ c & r & z \end{array} |$

LHS $= | \begin{array}{l} b + c & q + r & y + z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{array} |$

$= | \begin{matrix} 2 c & 2 r & 2 z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{matrix} | [R_{1} \to R_{1} + R_{2} - R_{3}]$

$= 2 | \begin{matrix} c & r & z \\ c + a & r + p & z + x \\ a + b & p + q & x + y \end{matrix} |$

$R_{1}$ से $2$ उभयनिष्ट लेने पर]

$= 2 | \begin{matrix} c & r & z \\ a & p & x \\ a + b & p + q & x + y \end{matrix} | [R_{2} \to R_{2} - R_{1}]$

$= 2 | \begin{array}{l} c & r & z \\ a & p & x \\ b & q & y \end{array} | [R_{3} \to R_{3} - R_{2}]$

$= 2 | \begin{array}{l} a & p & x \\ c & r & z \\ b & q & y \end{array} | [R_{1} \leftrightarrow R_{2}]$

$= 2 | \begin{array}{l} a & p & x \\ b & q & z \\ c & r & y \end{array} | [R_{2} \leftrightarrow R_{3}]$

= RHS

सारणिकों के गुणधर्मों का प्रयोग करके प्रश्न 6 से 14 तक को सिद्ध कीजिए।

6. $| \begin{matrix} 0 & a & - b \\ - a & 0 & - c \\ b & c & 0 \end{matrix} | = 0$

उत्तर: $| \begin{matrix} 0 & a & - b \\ - a & 0 & - c \\ b & c & 0 \end{matrix} |$

LHS = $| \begin{matrix} 0 & a & - b \\ - a & 0 & - c \\ b & c & 0 \end{matrix} |$

$= | \begin{matrix} 0 & a & - b \\ - a b & 0 & - b c \\ a b & a c & 0 \end{matrix} | [R_{2} \to b R_{2}, R_{3} \to a R_{3}]$

$= | \begin{matrix} 0 & a & - b \\ 0 & a c & - b c \\ a b & a c & 0 \end{matrix} | [R_{2} \to R_{2} + 3]$

$= a b (- a b c + a b c)$

$= a b (0)$

$= 0$

= RHS

7. $| \begin{matrix} - a^{2} & a b & a c \\ b a & - b^{2} & b c \\ c a & c b & - c^{2} \end{matrix} | = 4 a^{2} b^{2} c^{2}$

उत्तर: LHS = $| \begin{matrix} - a^{2} & a b & a c \\ b a & - b^{2} & b c \\ c a & c b & - c^{2} \end{matrix} |$

$= a b c | \begin{matrix} - a & a & a \\ b & - b & b \\ c & c & - c \end{matrix} |$

$C_{1}, C_{2}, C_{3}$ से $a, b, c$ उभयनिष्ट लेने पर]

$= a^{2} b^{2} c^{2} | \begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix} |$ $R_{1}, R_{2}, R_{3}$ से $a, b, c$ उभयनिष्ट लेने पर]

$= a^{2} b^{2} c^{2} | \begin{matrix} 0 & 1 & 1 \\ 0 & - 1 & 1 \\ 2 & 1 & - 1 \end{matrix} | [C_{1} \to C_{1} + C_{2}]$

$= a^{2} b^{2} c^{2} {2 (1 + 1)}$ $C_{1}$ के अनुदिश प्रसरण करने पर]

$= 4 a^{2} b^{2} c^{2}$

= RHS

8. (i) $| \begin{array}{l} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = (a - b) (b - c) (c - a)$

उत्तर: LHS = $| \begin{array}{l} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} |$

$= | \begin{matrix} 0 & a - b & a^{2} - b^{2} \\ 0 & b - c & b^{2} - c^{2} \\ 1 & c & c^{2} \end{matrix} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$= (a - b) (b - c) | \begin{matrix} 0 & 1 & a + b \\ 0 & 1 & b + c \\ 1 & c & c^{2} \end{matrix} |$ $R_{1}$ 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) $| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} | = (a - b) (b - c) (c - a) (a + b + c)$

उत्तर: LHS = $| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} |$

$= | \begin{matrix} 0 & 0 & 1 \\ a - b & b - c & c \\ a^{3} - b^{3} & b^{3} - c^{3} & c^{3} \end{matrix} | [C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3}]$

$= (a - b) (b - c) | \begin{matrix} 0 & 0 & 1 \\ 1 & 1 & c \\ a^{2} + a b + b^{2} & b^{2} + b c + c^{2} & c^{3} \end{matrix} |$ $C_{1}$ से $a - b$ , $C_{2}$ से $b - c$ उभयनिष्ट लेने पर]

$= (a - b) (b - c) {1 (b^{2} + b c + c^{2}) - (a^{2} + a b + b^{2})}$ $R_{1}$ के अनुदिश प्रसरण करने पर]

$= (a - b) (b - c) (c - a) (a + b + c)$

= RHS

9. $| \begin{array}{l} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array} | = (x - y) (y - z) (z - x) (x y + y z + z x)$

उत्तर: LHS = $| \begin{array}{l} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array} |$

$= | \begin{array}{l} x^{2} & x^{3} & x y z \\ y^{2} & y^{3} & y z x \\ z^{2} & z^{3} & z x y \end{array} | [R_{1} \to x R_{1}, R_{2} \to y R_{2}, R_{3} \to y R_{3}]$

$= x y z | \begin{array}{l} x^{2} & x^{3} & 1 \\ y^{2} & y^{3} & 1 \\ z^{2} & z^{3} & 1 \end{array} |$ $C_{3}$ से $x y z$ के उभयनिष्ट लेने पर]

$= x y z | \begin{matrix} x^{2} - y^{2} & x^{3} - y^{3} & 0 \\ y^{2} - z^{2} & y^{3} - z^{3} & 0 \\ z^{2} & z^{2} & 1 \end{matrix} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$= x y z (x - y) (y - z) | \begin{matrix} x + y & x^{2} + x y + y^{2} & 0 \\ y + z & y^{2} + y z + z^{2} & 0 \\ z^{2} & z^{2} & 1 \end{matrix} | [R_{1} \to x - y, R_{2} \to y - z]$

$= x y z (x - y) (y - z) {(x + y) (y^{2} + y z + z^{2}) - (y + z) (x^{2} + x y + y^{2})}$ $C_{3}$ के अनुदिश प्रसरण करने पर]

$= (x - y) (y - z) (z - x) (x y + y z + z x)$

= RHS

10. (i) $| \begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} | = (5 x + 4) (4 - x)^{2}$

उत्तर: LHS = $| \begin{matrix} x + 4 & 2 x & 2 x \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} |$

$R_{1} \to R_{1} + R_{2} + R_{3}$

= $| \begin{matrix} 5 x + 4 & 5 x + 4 & 5 x + 4 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} |$

$(5 x + 4) | \begin{matrix} 1 & 1 & 1 \\ 2 x & x + 4 & 2 x \\ 2 x & 2 x & x + 4 \end{matrix} |$ $C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}$

$= (5 x + 4) | \begin{matrix} 1 & 0 & 0 \\ 2 x & - x + 4 & 0 \\ 2 x & 0 & - x + 4 \end{matrix} |$

$= (5 x + 4) (4 - x) (4 - x) | \begin{matrix} 1 & 0 & 0 \\ 2 x & 1 & 0 \\ 2 x & 0 & 1 \end{matrix} |$

$= (5 x + 4) (4 - x)^{2} | \begin{matrix} 1 & 0 \\ 2 x & 1 \end{matrix} |$

$= (5 x + 4) (4 - x)^{2}$

= RHS

(ii) $| \begin{matrix} y + k & y & y \\ y & y + k & y \\ y & y & y + k \end{matrix} | = k^{2} (3 y + k)$

उत्तर: LHS = $| \begin{matrix} y + k & y & y \\ y & y + k & y \\ y & y & y + k \end{matrix} |$

$\begin{array}{r} R_{1} \to R_{1} + R_{2} + R_{3} \\ = | \begin{array}{c} 3 y + k & 3 y + k & 3 y + k \\ y & y + k & y \\ y & y & y + k \end{array} | \end{array}$

$= (3 y + k) | \begin{matrix} 1 & 1 & 1 \\ y & y + k & y \\ y & y & y + k \end{matrix} |$

$\begin{array}{r} C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} \\ = (3 y + k) | \begin{array}{l} 1 & 0 & 0 \\ y & k & 0 \\ y & 0 & k \end{array} | \end{array}$

$\begin{array}{r} = (3 y + k) k^{2} | \begin{array}{l} 1 & 0 & 0 \\ y & 1 & 0 \\ y & 0 & 1 \end{array} | \\ = (3 y + k) k^{2} | \begin{array}{l} 1 & 0 \\ y & 1 \end{array} | \\ = (3 y + k) k^{2} \end{array}$

= RHS

11. (i) $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | = (a + b + c)^{3}$

उत्तर: LHS = $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

$\begin{array}{r} R_{1} \to R_{1} + R_{2} + R_{3} \\ = | \begin{array}{c} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \\ = (a + b + c) | \begin{array}{c} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \\ C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} \end{array}$

$\begin{array}{r} = (a + b + c) | \begin{array}{c} 1 & 0 & 0 \\ 2 b & - (a + b + c) & 0 \\ 2 c & 0 & - (a + b + c) \end{array} | \\ = {(a + b + c)}^{3} | \begin{array}{c} 1 & 0 & 0 \\ 2 b & - 1 & 0 \\ 2 c & 0 & - 1 \end{array} | \\ = {(a + b + c)}^{3} (- 1) \times (- 1) = (a + b + c)^{3} \end{array}$

= RHS

(ii) $| \begin{matrix} x + y + 2 z & x & y \\ z & y + z + 2 x & y \\ z & x & z + x + 2 y \end{matrix} | = 2 (x + y + z)^{3}$

उत्तर: LHS = $| \begin{matrix} x + y + 2 z & x & y \\ z & y + z + 2 x & y \\ z & x & z + x + 2 y \end{matrix} |$

$\begin{array}{r} C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{array}{c} 2 (x + y + z) & x & y \\ 2 (x + y + z) & y + z + 2 x & y \\ 2 (x + y + z) & x & z + x + 2 y \end{array} | \\ = 2 (x + y + z) | \begin{array}{c} 1 & x & y \\ 1 & y + z + 2 x & y \\ 1 & x & z + x + 2 y \end{array} | \\ R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} \end{array}$

$\begin{array}{r} = 2 (x + y + z) | \begin{array}{c} 1 & x & y \\ 0 & y + z + x & 0 \\ 0 & 0 & z + x + y \end{array} | \\ = 2 (x + y + z)^{3} | \begin{array}{c} 1 & x & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} | \\ = 2 (x + y + z)^{3} \end{array}$

= RHS

12. $| \begin{matrix} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{matrix} | = {(1 + x^{3})}^{2}$

उत्तर: LHS = $| \begin{matrix} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{matrix} |$

$\begin{array}{r} R_{1} \to R_{1} + R_{2} + R_{3} \\ = | \begin{array}{c} 1 + x + x^{2} & 1 + x + x^{2} & 1 + x + x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array} | \end{array}$

$\begin{array}{r} = (1 + x + x^{2}) | \begin{array}{c} 1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array} | \\ C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1} \end{array}$

$\begin{array}{r} = (1 + x + x^{2}) | \begin{array}{c} 1 & 0 & 0 \\ x^{2} & 1 - x^{2} & x - x^{2} \\ x & x^{2} - x & 1 - x^{2} \end{array} | \\ = (1 + x + x^{2}) (1 - x) (1 - x) | \begin{array}{c} 1 & 0 & 0 \\ x^{2} & 1 + x & x \\ x & - x & 1 \end{array} | \\ = (1 - x^{3}) (1 - x) | \begin{array}{c} 1 & 0 & 0 \\ x^{2} & 1 + x & x \\ x & - x & 1 \end{array} | \\ = (1 - x^{3}) (1 - x) (1) | \begin{array}{c} 1 + x & x \\ - x & 1 \end{array} | \\ = (1 - x^{3}) (1 - x) (1 + x + x^{2}) \\ = {(1 - x^{3})}^{2} \end{array}$

= RHS

13. $| \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b \\ 2 a b & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} | = {(1 + a^{2} + b^{2})}^{3}$

उत्तर: LHS = $| \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b \\ 2 a b & 1 - a^{2} + b^{2} & 2 a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} |$

$R_{2} \to R_{2} - a R_{3}, R_{1} \to R_{1} + b R_{3}$

$= | \begin{matrix} 1 + a^{2} + b^{2} & 0 & - b (1 + a^{2} + b^{2}) \\ 0 & 1 + a^{2} + b^{2} & a (1 + a^{2} + b^{2}) \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} |$

$= {(1 + a^{2} + b^{2})}^{2} | \begin{matrix} 1 & 0 & - b \\ 0 & 1 & a \\ 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} |$

$= {(1 + a^{2} + b^{2})}^{2} (1 - a^{2} - b^{2} + 2 a^{2} - b (- 2 b))$

$\begin{array}{r} = {(1 + a^{2} + b^{2})}^{2} (1 + a^{2} + b^{2}) \\ = {(1 + a^{2} + b^{2})}^{3} \end{array}$

= RHS

14. $| \begin{matrix} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ c a & c b & c^{2} + 1 \end{matrix} | = 1 + a^{2} + b^{2} + c^{2}$

उत्तर: LHS = $| \begin{matrix} 1 + a^{2} & a b & a c \\ a b & 1 + b^{2} & b c \\ c a & c b & c^{2} + 1 \end{matrix} |$

$\begin{array}{r} = a b c | \begin{array}{c} a + \frac{1}{a} & b & c \\ a & b + \frac{1}{b} & c \\ a & b & c + \frac{1}{c} \end{array} | \\ R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1} \end{array}$

$\begin{array}{r} = a b c | \begin{array}{c} a + \frac{1}{a} & b & c \\ - \frac{1}{a} & \frac{1}{b} & 0 \\ - \frac{1}{a} & 0 & \frac{1}{c} \end{array} | \\ C_{1} \to a C_{1}, C_{2} \to b C_{2}, C_{3} \to c C_{3} \end{array}$

$\begin{array}{r} = a b c \times \frac{1}{a b c} | \begin{array}{c} 1 + a^{2} & b^{2} & c^{2} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} | \\ = - 1 | \begin{array}{c} b^{2} & c^{2} \\ 1 & 0 \end{array} | + 1 | \begin{array}{c} 1 + a^{2} & b^{2} \\ - 1 & 1 \end{array} | \\ = 1 + a^{2} + b^{2} + c^{2} \end{array}$

= RHS

प्रश्न संख्या 15 तथा 16 मे सही उत्तर चुनिये।

15. यदि $A$ एक $3 \times 3$ कोटी का वर्ग आव्यूह है तो $| k A |$ का मान होगा:

(a) $k | A |$

(b) $k^{2} | A |$

(d) $3 k | A |$

उत्तर: $| A | = | \begin{array}{l} a & b & c \\ d & e & f \\ g & h & i \end{array} |$

$\begin{array}{r} | k A | = | \begin{array}{l} k a & k b & k c \\ k d & k e & k f \\ k g & k h & k i \end{array} | \\ | k A | = k^{3} | \begin{array}{l} a & b & c \\ d & e & f \\ g & h & i \end{array} | \\ | k A | = k^{3} | A | \end{array}$

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।

(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 $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ से होकर जाने वाले त्रिभुज का क्षेत्रफल,

$Δ = \frac{1}{2} | \begin{array}{l} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$

अभीष्ट त्रिभुज का क्षेत्रफल

$\begin{array}{r} Δ = \frac{1}{2} | \begin{array}{l} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{array} | \\ = \frac{1}{2} [1 (0 - 3) - 0 (6 - 4) + 1 (18 - 0)] \\ = \frac{1}{2} [- 3 + 18] = \frac{15}{2} \end{array}$

(ii) $(2, 7), (1, 2), (10, 8)$

उत्तर: अभीष्ट त्रिभुज का क्षेत्रफल

$Δ = \frac{1}{2} | \begin{matrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{matrix} |$

$\begin{array}{r} = \frac{1}{2} [2 (1 - 8) - 7 (1 - 10) + 8 (8 - 10)] \\ = \frac{1}{2} [- 16 + 63] \end{array}$

$= \frac{47}{2}$

(iii) $(- 2, - 3), (3, 2), (- 1, - 8)$

उत्तर: अभीष्ट त्रिभुज का क्षेत्रफल

$Δ = \frac{1}{2} | \begin{matrix} - 2 & - 3 & 1 \\ 3 & 2 & 1 \\ - 1 & - 8 & 1 \end{matrix} |$

$\begin{array}{r} = \frac{1}{2} [- 2 (2 + 8) + 3 (3 + 1) + (- 24 + 2)] \\ = \frac{1}{2} [- 2 (10) + 3 (4) + (- 22)] \\ = \frac{1}{2} [- 20 + 12 - 22] \\ = - \frac{30}{2} = - 15 \end{array}$

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 $A (a, b + c), B (b . c + a), C (c, a + b)$

$Δ$ का क्षेत्रफल = $Δ = \frac{1}{2} | \begin{array}{l} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$

जहा $x_{1} = a, y_{1} = b + c, x_{2} = b, y_{2} = c + a, x_{3} = c, y_{3} = a + b$

$= \frac{1}{2} | \begin{array}{l} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{array} | (C_{1} \to C_{1} + C_{2})$

$\begin{array}{r} = \frac{1}{2} | \begin{array}{l} a + b + c & b + c & 1 \\ a + b + c & c + a & 1 \\ a + b + c & a + b & 1 \end{array} | \\ = \frac{1}{2} (a + b + c) | \begin{array}{l} 1 & b + c & 1 \\ 1 & c + a & 1 \\ 1 & a + b & 1 \end{array} | \\ = \frac{1}{2} (a + b + c) \times 0 \end{array}$

$Δ$ 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 = $= \frac{1}{2} | \begin{array}{l} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$

$\begin{array}{r} \frac{1}{2} | \begin{array}{l} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array} | = \pm 4 \\ | \begin{array}{l} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array} | = \pm 8 \\ (- 2) | \begin{array}{l} k & 1 \\ 4 & 1 \end{array} | = \pm 8 \\ k - 4 = \pm 4 \\ k = 0, 8 \end{array}$

(ii) $(- 2, 0), (0, 4), (0, k)$

उत्तर: त्रिभुज का क्षेत्रफल = $= \frac{1}{2} | \begin{array}{l} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$

$\begin{array}{r} = \frac{1}{2} | \begin{array}{c} - 2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{array} | \\ = \frac{1}{2} [- 2 (4 - k) - 0 (0 - 0) + 1 (0 - 0) \\ = \frac{1}{2} \times (- 2) (4 - k) = k - 4 = \pm 4 \\ + v e, k - 4 = 4; k = 8 \\ - v e, k - 4 = 4; k = 0 \end{array}$

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।

$Δ = \frac{1}{2} | \begin{array}{l} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$

जहा $x_{1} = x, y_{1} = y, x_{2} = 1, y_{2} = 2, x_{3} = 3, y_{3} = 6$

$\begin{array}{r} = \frac{1}{2} | \begin{array}{c} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{array} | \\ = \frac{1}{2} [x (2 - 6) - y (1 - 3) + 1 (6 - 6)] \\ = \frac{1}{2} [x \times (- 4) - y (- 2) + 1 \times 0] \\ = \frac{1}{2} [- 4 x + 2 y] \end{array}$

$\begin{array}{r} = \frac{1}{2} \times 2 (- 2 x + y) \\ = - 2 x + y \end{array}$

बिन्दु संरेख है।

इसलिए $Δ$ त्रिभुज का क्षेत्रफल शून्य होगा।

$\begin{array}{r} 0 = - 2 x + y \\ 2 x - y = 0 \end{array}$

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।

$\begin{array}{r} A P B = 0 \\ \frac{1}{2} | \begin{array}{l} 3 & 1 & ∣ 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{array} | = 0 \\ | \begin{array}{l} 3 & 1 & 1 \\ 9 & 3 & 1 \\ x & y & 1 \end{array} | = 0 \end{array}$

5. यदि शीर्ष $(2, - 6), (5, 4), (k, 4)$ वाले त्रिभुज का क्षेत्रफल $35$ वर्ग इकाई हो तो का मान है।

(a) $12$

(b) $- 2$

(d) $12, - 2$

उत्तर: दिया है, त्रिभुज के शीर्ष $(2, - 6), (5, 4), (k, 4)$

$Δ = \frac{1}{2} | \begin{array}{l} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} |$

जहा $x_{1} = 2, y_{1} = 6, x_{2} = 5, y_{2} = 4, x_{3} = k, y_{3} = 4$

ज्ञात $Δ$ का क्षेत्रफल = $\pm 35$

$\begin{array}{r} \pm 35 = \frac{1}{2} [2 (4 - 4) + 6 (5 - k) + k (120 - 4 k)] \\ \pm 35 = \frac{1}{2} [2 \times 0 + 6 (5 - k) + 1 (20 - 4 k)] \\ \pm 70 = 6 (5 - k) + 20 - 4 k \\ \pm 70 = 30 - 6 k + 20 - 4 k \\ \pm 70 = 50 - 10 k \\ \pm 70 = 5 - k \\ + v e, 7 = 5 - k; k = 5 - 7 = - 2 \\ - v e, - 7 = 5 - k; - 12 = - k; k = 12 \end{array}$

अतः $k = 12, - 2$

अतः विकल्प (d) सही है।

प्रश्नावली 4.4

निम्नलिखित सारणिकों के अवयवों के उपसारणिक एवं सहखंड लिखिए:

1. (i) $| \begin{matrix} 2 & - 4 \\ 0 & 3 \end{matrix} |$

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) $| \begin{array}{l} a & c \\ b & d \end{array} |$

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) $| \begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} |$

उत्तर: उपसारणिक

$\begin{array}{r} M_{11} = | \begin{array}{l} 1 & 0 \\ 0 & 1 \end{array} | = 1 - 0 = 1, M_{12} = | \begin{array}{l} 0 & 0 \\ 0 & 1 \end{array} | = 0 \\ M_{13} = | \begin{array}{l} 0 & 1 \\ 0 & 0 \end{array} | = 0, M_{21} = | \begin{array}{l} 0 & 0 \\ 0 & 1 \end{array} | = 0 \\ M_{22} = | \begin{array}{l} 1 & 0 \\ 0 & 1 \end{array} | = 1, M_{23} = | \begin{array}{l} 1 & 0 \\ 0 & 0 \end{array} | = 0 \\ M_{31} = | \begin{array}{l} 0 & 0 \\ 0 & 1 \end{array} | = 0, M_{32} = | \begin{array}{l} 1 & 0 \\ 0 & 0 \end{array} | = 0 \\ M_{31} = | \begin{array}{l} 1 & 0 \\ 0 & 1 \end{array} | = 0 \end{array}$

तथा सहखंड

$\begin{array}{r} A_{11} = (- 1)^{1 + 1} M_{11} = 1, A_{12} = (- 1)^{1 + 2} M_{12} = (- 1) \times 0 = 0 \\ A_{13} = (- 1)^{1 + 3} M_{12} = 1 \times 0 = 0, A_{21} = (- 1)^{2 + 1} M_{21} = 0 \end{array}$

$\begin{array}{r} A_{22} = (- 1)^{2 + 2} M_{22} = 1, A_{23} = (- 1)^{2 + 3} M_{23} = 0 \\ A_{31} = (- 1)^{3 + 1} M_{31} = 0, A_{32} = (- 1)^{3 + 2} M_{32} = 0 \\ A_{33} = (- 1)^{3 + 3} M_{33} = 1 .1 = 1 \end{array}$

(ii) $| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & - 1 \\ 0 & 1 & 2 \end{matrix} |$

उत्तर: उपसारणिक

$\begin{array}{r} M_{11} = | \begin{array}{c} 5 & - 1 \\ 1 & 2 \end{array} | = 5 \times 2 - 1 (1) = 11, M_{12} = | \begin{array}{c} 3 & - 1 \\ 0 & 2 \end{array} | = 3 \times 2 - 0 = 6 \\ M_{13} = | \begin{array}{l} 3 & 5 \\ 0 & 1 \end{array} | = 3 \times 1 - 0 = 3, M_{21} = | \begin{array}{l} 0 & 4 \\ 1 & 2 \end{array} | = 0 \times 2 - 1 \times 4 = - 4 \\ M_{22} = | \begin{array}{l} 1 & 4 \\ 0 & 2 \end{array} | = 1 \times 2 - 0 = 2, M_{23} = | \begin{array}{l} 1 & 0 \\ 0 & 2 \end{array} | = 1 \times 1 = 1 \\ M_{32} = | \begin{array}{c} 1 & 4 \\ 3 & - 1 \end{array} | = 1 \times - (1) - 3 \times 4 = 13, M_{33} = | \begin{array}{l} 1 & 0 \\ 3 & 5 \end{array} | = 5 - 0 = 5 \end{array}$

तथा सहखंड

$\begin{array}{r} A_{11} = (- 1)^{1 + 1} M_{11} = 11, A_{12} = (- 1)^{1 + 2} M_{12} = (- 1) \times 6 = - 6 \\ A_{13} = (- 1)^{1 + 3} M_{12} = 1 .3 = 3, A_{21} = (- 1)^{2 + 1} M_{21} = (- 1) \times (- 4) = 4 \end{array}$ $\begin{array}{r} A_{22} = (- 1)^{2 + 2} M_{22} = 1 \times 2 = 2, A_{23} = (- 1)^{2 + 3} M_{23} = (- 1) 1 = - 1 \\ A_{31} = (- 1)^{3 + 1} M_{31} = 1 . (- 20) = - 20, A_{32} = (- 1)^{3 + 2} M_{32} = (- 1) (- 13) = 13 \\ A_{33} = (- 1)^{3 + 3} M_{33} = 1 .5 = 1 \end{array}$

3. दूसरी पंक्ति के अवयवों के सहखंडों का प्रयोग करके का मान ज्ञात कीजिए।

उत्तर: $A_{21} = (- 1)^{2 + 1} | \begin{array}{l} 3 & 8 \\ 2 & 3 \end{array} | = (- 1) \times [3 \times 3 - 2 \times 8] = 7$

$\begin{array}{r} A_{22} = (- 1)^{2 + 2} | \begin{array}{l} 5 & 8 \\ 1 & 3 \end{array} | = 1 [5 \times 3 - 2 \times 8] = 7 \\ A_{23} = (- 1)^{2 + 3} | \begin{array}{l} 5 & 3 \\ 1 & 2 \end{array} | = (- 1) [5 \times 2 - 3 \times 1] = - 7 \\ Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} \\ = 2 \times 7 + 0 \times 7 + 1 \times (- 7) \\ = 14 - 7 = 7 \end{array}$

4. तीसरे स्तम्भ के अवयवों के सहखंडों का प्रयोग करके $Δ = | \begin{array}{l} 1 & x & y z \\ 1 & y & z x \\ 1 & 2 & x y \end{array} |$ का मान ज्ञात कीजिए।

उत्तर: $A_{13} = - 1^{1 + 3} | \begin{matrix} 1 & y \\ 1 & z \end{matrix} | = (1) (z - y) = (z - y)$

$A_{23} = - 1^{2 + 3} | \begin{array}{l} 1 & x \\ 1 & z \end{array} | = (1) (z - x) = - (x - z)$

$\begin{array}{r} A_{33} = - 1^{1 + 3} | \begin{array}{l} 1 & x \\ 1 & y \end{array} | = (1) (y - x) = (y - x) \\ Δ = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33} \\ = y z (z - y) + z x (x - z) + x y (y - x) \\ = y z^{2} - y^{2} z + z x^{2} - z^{2} x + x y^{2} - x^{2} y \\ = z x^{2} - x^{2} y + x y^{2} - z^{2} y + y z^{2} - y^{2} z \\ = x^{2} ((z - y) + z (y - z) (y + z) + y z (z - y) \\ = (z - y) [x^{2} - x (y + z) + y z] \\ = (z - y) [x^{2} - x y - x z + y z] \\ = (z - y) [x (x - y) - z (x - y)] \\ = (z - y) (x - y) (x - z) \\ = (x - y) (y - z) (z - x) \end{array}$

5. यदि $Δ = | \begin{array}{l} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} |$ और $a_{y}$ का सहखंड $A_{y}$ हो तो $Δ$ 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) $a_{11} \times A_{11} + a_{12} A_{21} + a_{13} A_{31}$

(d) $a_{11} \times A_{11} + a_{21} A_{21} + a_{31} A_{31}$

उत्तर: $Δ$ = 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. $[\begin{array}{r} 1 2 \\ 3 4 \end{array}]$

उत्तर: यहा $A = [\begin{array}{l} 1 & 2 \\ 3 & 4 \end{array}]$

इसलिए $a d j A = [\begin{matrix} 4 & - 2 \\ - 3 & 1 \end{matrix}]$

2. $[\begin{array}{r} 1 - 1 2 \\ 2 3 5 \\ - 2 0 1 \end{array}]$

उत्तर: यहा $A = [\begin{matrix} 1 & - 1 & 2 \\ 2 & 3 & 5 \\ - 2 & 0 & 1 \end{matrix}]$

इसलिए $a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{matrix} 3 & 1 & - 11 \\ - 12 & 5 & - 1 \\ 6 & 2 & 5 \end{matrix}]$

प्रश्न 3 और 4 मे सत्यापित कीजिए की $A (a d j A) = (a d j A) \times A = | A | . I$ है।

3. $[\begin{array}{r} 2 3 \\ - 4 - 6 \end{array}]$

उत्तर: यहा $A = [\begin{matrix} 2 & 3 \\ - 4 & - 6 \end{matrix}]$

इसलिए $a d j A = [\begin{matrix} - 6 & - 3 \\ 4 & 2 \end{matrix}]$

$\begin{array}{r} A (a d j A) = [\begin{array}{c} 2 & 3 \\ - 4 & - 6 \end{array}] [\begin{array}{c} - 6 & - 3 \\ 4 & 2 \end{array}] \\ = [\begin{array}{c} - 12 + 12 & - 6 + 6 \\ 24 - 24 & 12 - 12 \end{array}] = [\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] \end{array}$

$\begin{array}{r} (a d j A) A = [\begin{array}{c} - 6 & - 3 \\ 4 & 2 \end{array}] [\begin{array}{c} 2 & 3 \\ - 4 & - 6 \end{array}] \\ = [\begin{array}{c} - 12 + 12 & - 18 + 18 \\ 8 - 8 & 12 - 12 \end{array}] = [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}] \end{array}$

$\begin{array}{r} | A | I = 0 \times [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}] \\ A (a d j A) = (a d j A) A = | A | I = [\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] \end{array}$

4. $[\begin{array}{r} 1 - 1 2 \\ 3 0 - 2 \\ 1 0 3 \end{array}]$

उत्तर: $A = [\begin{matrix} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{matrix}]$

$\begin{array}{r} a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{array}{c} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{array}] \\ A (a d j A) = [\begin{array}{c} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{array}] [\begin{array}{c} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{array}] \\ = [\begin{array}{c} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}] \\ (a d j A) A = [\begin{array}{c} 0 & 3 & 2 \\ - 11 & 1 & 8 \\ 0 & - 1 & 3 \end{array}] [\begin{array}{c} 1 & - 1 & 2 \\ 3 & 0 & - 2 \\ 1 & 0 & 3 \end{array}] = [\begin{array}{c} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}] \end{array}$

$\begin{array}{r} | A | I = 11 [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{c} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}] \\ A (a d j A) = (a d j A) A = | A | I = [\begin{array}{c} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}] \end{array}$

प्रश्न 5 से 11 मे दिए गए प्रत्येक आवयहो के व्युत्क्रम ज्ञात कीजिए।

5. $[\begin{array}{r} 2 - 2 \\ 4 3 \end{array}]$

उत्तर: यहा $A = [\begin{matrix} 2 & - 2 \\ 4 & 3 \end{matrix}]$

इसलिए $a d j A = [\begin{matrix} 3 & 2 \\ - 4 & 2 \end{matrix}]$

$| A | = 6 + 8 = 14 \neq 0 \to A^{- 1}$ का अस्तित्व है।

$A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{14} [\begin{matrix} 3 & 2 \\ - 4 & 2 \end{matrix}]$

6. $[\begin{array}{r} - 1 5 \\ - 3 2 \end{array}]$

उत्तर: यहा $A = [\begin{array}{l} - 1 & 5 \\ - 3 & 2 \end{array}]$

इसलिए $a d j A = [\begin{array}{l} 2 & - 5 \\ 3 & - 1 \end{array}]$

$| A | = - 2 + 15 = 13 \neq 0 \to A^{- 1}$ का अस्तित्व है।

$A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{13} [\begin{array}{l} 2 & - 5 \\ 3 & - 1 \end{array}]$

7. $[\begin{array}{r} 1 2 3 \\ 0 2 4 \\ 0 0 5 \end{array}]$

उत्तर: यहा $A = [\begin{array}{l} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}]$

इसलिए $a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{matrix} 10 & - 10 & 2 \\ 0 & 5 & - 4 \\ 0 & 0 & 2 = \end{matrix}]$

$| A | = 1 (10 - 0) - 2 (0 - 0) + 3 (0 - 0) = 10 \neq 0 \to A^{- 1}$ का अस्तित्व है।

8. $[\begin{array}{r} 1 0 0 \\ 3 3 0 \\ 5 2 - 1 \end{array}]$

उत्तर: यहा $A = [\begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & - 1 \end{matrix}]$

इसलिए $a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{matrix} - 3 & 0 & 0 \\ 3 & - 1 & 0 \\ - 9 & - 2 & 3 \end{matrix}]$

$| A | = 1 (- 3 - 0) - 0 (- 3 - 0) + 0 (6 - 15) = - 3 \neq 0 \to A^{- 1}$ का अस्तित्व है।

9. $[\begin{array}{r} 2 1 3 \\ 4 - 1 0 \\ - 7 2 1 \end{array}]$

उत्तर: यह $A = [\begin{matrix} 2 & 1 & 3 \\ 4 & - 1 & 0 \\ - 7 & 2 & 1 \end{matrix}]$

इसलिए $a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{matrix} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{matrix}]$

$| A | = 2 (- 1 - 0) - 1 (4 - 0) + 3 (8 - 7) = - 3 \neq 0 \to A^{- 1}$ का अस्तित्व है।

$\begin{array}{r} A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{| A |} [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \\ = \frac{1}{- 3} [\begin{array}{c} - 1 & 5 & 3 \\ - 4 & 23 & 12 \\ 1 & - 11 & - 6 \end{array}] \end{array}$

10. $[\begin{array}{r} 1 - 1 2 \\ 0 2 - 3 \\ 3 - 2 4 \end{array}]$

उत्तर: यहा $A = [\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}]$

इसलिए $a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{matrix} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{matrix}]$

$| A | = 1 (8 - 6) + 1 (0 + 9) + 2 (0 - 6) = - 1 \neq 0 \to A^{- 1}$ का अस्तित्व है।

$\begin{array}{r} A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{| A |} [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \\ = \frac{1}{- 1} [\begin{array}{c} 2 & 0 & - 1 \\ - 9 & - 2 & 3 \\ - 6 & - 1 & 2 \end{array}] = [\begin{array}{c} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{array}] \end{array}$

11. $[\begin{array}{r} 1 0 0 \\ 0 c o s α s i n α \\ 0 s i n α c o s α \end{array}]$

उत्तर: यहा $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & c o s α & s i n α \\ 0 & s i n α & - c o s α \end{matrix}]$

इसलिए $a d j A = [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & - c o s α & - s i n α \\ 0 & - s i n α & c o s α \end{matrix}]$

$| A | = 1 (- c o s^{2} α - s i n^{2} α) + 0 (0 - 0) + 0 (0 - 0) = - 1 \neq 0 \to A^{- 1}$ का अस्तित्व है।

$\begin{array}{r} A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{| A |} [\begin{array}{l} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \\ = \frac{1}{- 1} [\begin{array}{c} 1 & 0 & 0 \\ 0 & - c o s α & - s i n α \\ 0 & - s i n α & c o s α \end{array}] = [\begin{array}{c} - 1 & 0 & 0 \\ 0 & c o s α & s i n α \\ 0 & s i n α & - c o s α \end{array}] \end{array}$

12. यदि $A = [\begin{array}{r} 3 7 \\ 2 5 \end{array}]$ और $B = [\begin{array}{r} 6 8 \\ 7 9 \end{array}]$ है तो सत्यापित कीजिए की ${(A B)}^{- 1} = B^{- 1} A^{- 1}$

उत्तर: यहा $A = [\begin{array}{l} 3 & 7 \\ 2 & 5 \end{array}]$

$| A | = 15 - 14 = 1 \neq 0 \to A^{- 1}$ का अस्तित्व है।

$A^{- 1} = \frac{1}{| A |} (a d j A) = \frac{1}{1} [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}] = [\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}]$

यहा $B = [\begin{array}{l} 6 & 8 \\ 7 & 9 \end{array}]$

$| B | = 54 - 56 = - 2 \neq 0 \to B^{- 1}$ का अस्तित्व है।

$B^{- 1} = \frac{1}{| B |} (a d j B) = \frac{1}{- 2} [\begin{matrix} 9 & - 8 \\ - 7 & 6 \end{matrix}] = [\begin{matrix} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{matrix}]$

$\begin{array}{r} B^{- 1} A^{- 1} = [\begin{array}{c} - \frac{9}{2} & 4 \\ \frac{7}{2} & - 3 \end{array}] [\begin{array}{c} 5 & - 7 \\ - 2 & 3 \end{array}] = [\begin{array}{c} - \frac{45}{2} - 8 & \frac{63}{2} + 12 \\ \frac{35}{2} + 6 & - \frac{49}{2} - 9 \end{array}] \\ = [\begin{array}{c} - \frac{81}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array}] \\ A B = [\begin{array}{l} 3 & 7 \\ 2 & 5 \end{array}] [\begin{array}{l} 6 & 8 \\ 7 & 9 \end{array}] = [\begin{array}{l} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{array}] = [\begin{array}{l} 67 & 87 \\ 47 & 61 \end{array}] \end{array}$

$| A B | = 4087 - 4089 = - 2 \neq 0 \to (A B)^{- 1}$ का अस्तित्व है।

$\begin{array}{r} {(A B)}^{- 1} = \frac{1}{| A B |} (a d j A B) = \frac{1}{- 2} [\begin{array}{c} 61 & - 87 \\ - 47 & 67 \end{array}] = [\begin{array}{c} - \frac{61}{2} & \frac{87}{2} \\ \frac{47}{2} & - \frac{67}{2} \end{array}] \\ {(A B)}^{- 1} = B^{- 1} A^{- 1} \end{array}$

13. यदि $A = [\begin{array}{r} 3 1 \\ - 1 2 \end{array}]$ है तो दर्शाइए की $A^{2} - 5 A + 7 I = 0$ है इसकी सहायता से $A^{- 1}$ ज्ञात कीजिए।

उत्तर: LHS = $A^{2} - 5 A + 7 I = A A - 5 A + 7 I$

$\begin{array}{r} = [\begin{array}{c} 3 & 1 \\ - 1 & 2 \end{array}] [\begin{array}{c} 3 & 1 \\ - 1 & 2 \end{array}] - 5 [\begin{array}{c} 3 & 1 \\ - 1 & 2 \end{array}] + 7 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] \\ = [\begin{array}{c} 9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4 \end{array}] - [\begin{array}{c} 15 & 5 \\ - 5 & 10 \end{array}] + [\begin{array}{c} 7 & 0 \\ 0 & 7 \end{array}] \\ = [\begin{array}{c} 8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7 \end{array}] = [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}] \\ = 0 \end{array}$

= RHS

$\begin{array}{r} A^{2} - 5 A + 7 I = 0 \\ A^{2} - 5 A = - 7 I \\ A A A^{- 1} - 5 A A^{- 1} = - 7 I A^{- 1} \\ 7 A^{- 1} = 5 I - A I = 5 [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] - [\begin{array}{c} 3 & 1 \\ - 1 & 2 \end{array}] \\ = [\begin{array}{l} 5 & 0 \\ 0 & 5 \end{array}] - [\begin{array}{c} 3 & 1 \\ - 1 & 2 \end{array}] = [\begin{array}{c} 2 & - 1 \\ 1 & 3 \end{array}] \\ A^{- 1} = \frac{1}{7} [\begin{array}{c} 2 & - 1 \\ 1 & 3 \end{array}] \end{array}$

14. आव्यूह $A = [\begin{array}{r} 3 2 \\ 1 1 \end{array}]$ के लिए और ऐसी संख्या ज्ञात कीजिए ताकि $A^{2} + a A + b I = 0$ हो।

उत्तर: $A^{2} + a A + b I = 0$

$\begin{array}{r} = [\begin{array}{l} 3 & 2 \\ 1 & 1 \end{array}] [\begin{array}{l} 3 & 2 \\ 1 & 1 \end{array}] + a [\begin{array}{l} 3 & 2 \\ 1 & 1 \end{array}] + b [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}] \\ = [\begin{array}{l} 9 + 2 & 6 + 2 \\ 3 + 1 & 2 + 1 \end{array}] - [\begin{array}{c} 3 a & 2 a \\ a & a \end{array}] + [\begin{array}{c} b & 0 \\ 0 & b \end{array}] = [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}] \\ = [\begin{array}{l} 11 + 3 a + b & 8 + 2 a + 0 \\ 4 + a + 0 & 3 + a + b \end{array}] = [\begin{array}{l} 0 & 0 \\ 0 & 0 \end{array}] \\ 4 + a = 0 \\ a = - 4 \\ 3 + a + b = 0 \\ 3 - 4 + b = 0 \\ b = 1 \\ a = - 4, b = 1 \end{array}$

15. आव्यूह $A = [\begin{array}{r} 1 1 1 \\ 1 2 - 3 \\ 2 - 1 3 \end{array}]$ के लिए दर्शाइए की $A^{3} - 6 A^{2} + 5 A + 11 I = 0$ है। इसकी सहायता से $A^{- 1}$ ज्ञात कीजिए।

उत्तर: $A = [\begin{array}{r} 1 1 1 \\ 1 2 - 3 \\ 2 - 1 3 \end{array}]$

$\begin{array}{r} A^{2} = [\begin{array}{c} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] = [\begin{array}{c} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] \\ A^{3} = A^{2} A = [\begin{array}{c} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] \\ = [\begin{array}{c} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}] \end{array}$

LHS = $A^{3} - 6 A^{2} + 5 A + 11 I = 0$

$\begin{array}{r} = [\begin{array}{c} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}] - 6 [\begin{array}{c} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{array}] + 5 [\begin{array}{c} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{array}] + 11 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ = [\begin{array}{c} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{array}] - [\begin{array}{c} 24 & 12 & 6 \\ - 18 & 48 & - 84 \\ 42 & - 18 & 84 \end{array}] + [\begin{array}{c} 5 & 5 & 5 \\ 5 & 10 & - 15 \\ 10 & - 5 & 15 \end{array}] + [\begin{array}{c} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{array}] \\ = [\begin{array}{c} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0 \\ - 23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 + 0 \\ 32 - 42 + 10 + 0 & - 13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{array}] \\ = [\begin{array}{l} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] = 0 \end{array}$

= RHS

$\begin{array}{r} A^{3} - 6 A^{2} + 5 A + 11 I = 0 \\ A^{3} - 6 A^{2} + 5 A = - 11 I \\ A^{2} A A^{- 1} - 6 A A A^{- 1} + 5 A A^{- 1} = - 11 I A^{- 1} \end{array}$

$11 A^{- 1} = - A^{2} + 6 A - 5 I = [\begin{matrix} - 4 & - 2 & - 1 \\ 3 & - 8 & 14 \\ - 7 & 3 & - 14 \end{matrix}] + 6 [\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3 \end{matrix}] - 5 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ $\begin{array}{r} = [\begin{array}{c} - 4 & - 2 & - 1 \\ 3 & - 8 & 14 \\ - 7 & 3 & - 14 \end{array}] + [\begin{array}{c} 6 & 6 & 6 \\ 6 & 12 & - 18 \\ 12 & - 6 & 18 \end{array}] - [\begin{array}{c} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}] \\ = [\begin{array}{c} - 3 & 4 & 5 \\ 9 & - 1 & - 4 \\ 5 & - 3 & - 1 \end{array}] \\ A^{- 1} = \frac{1}{11} [\begin{array}{c} - 3 & 4 & 5 \\ 9 & - 1 & - 4 \\ 5 & - 3 & - 1 \end{array}] \end{array}$

16. यदि $A = [\begin{array}{r} 2 - 1 1 \\ - 1 2 - 1 \\ 1 - 1 2 \end{array}]$ तो सत्यापित कीजिए की $A^{3} - 6 A^{2} + 9 A - 4 I = 0$ है तथा इसकी सहायता से $A^{- 1}$ ज्ञात कीजिए।

उत्तर: $A = [\begin{array}{r} 2 - 1 1 \\ - 1 2 - 1 \\ 1 - 1 2 \end{array}]$

$\begin{array}{r} A^{2} = [\begin{array}{c} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] [\begin{array}{c} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] = [\begin{array}{c} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] \\ A^{3} = A^{2} A = [\begin{array}{c} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] [\begin{array}{c} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] = [\begin{array}{c} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{array}] \end{array}$

LHS = $A^{3} - 6 A^{2} + 9 A - 4 I = 0$

$= [\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] - 6 [\begin{matrix} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{matrix}] + 9 [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] - 4 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$= [\begin{matrix} 22 & - 21 & 21 \\ - 21 & 22 & - 21 \\ 21 & - 21 & 22 \end{matrix}] - [\begin{matrix} 36 & - 30 & 30 \\ - 30 & 36 & - 30 \\ 30 & - 30 & 36 \end{matrix}] + [\begin{matrix} 18 & - 9 & 9 \\ - 9 & 18 & - 9 \\ 9 & - 9 & 18 \end{matrix}] - [\begin{array}{l} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}]$

$\begin{array}{r} = [\begin{array}{l} 22 - 36 + 18 - 4 & - 21 + 30 - 9 - 0 & - 21 - 30 + 9 - 0 \\ - 21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 - 0 \\ 21 - 30 + 9 - 0 & - 21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 \end{array}] \\ = [\begin{array}{l} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] = 0 \end{array}$

= RHS

$\begin{array}{r} A^{3} - 6 A^{2} + 9 A - 4 I = 0 \\ A^{3} - 6 A^{2} + 9 A = 4 I \\ A^{2} A A^{- 1} - 6 A A A^{- 1} + 9 A A^{- 1} = 4 I A^{- 1} \\ 4 A^{- 1} = A^{2} - 6 A + 9 I = [\begin{array}{c} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] - 6 [\begin{array}{c} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] + 9 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{array}$

$\begin{array}{r} = [\begin{array}{c} 6 & - 5 & 5 \\ - 5 & 6 & - 5 \\ 5 & - 5 & 6 \end{array}] + [\begin{array}{c} 12 & - 6 & 6 \\ - 6 & 12 & - 6 \\ 6 & - 6 & 12 \end{array}] - [\begin{array}{c} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}] \\ = [\begin{array}{c} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}] \\ A^{- 1} = \frac{1}{4} [\begin{array}{c} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}] \end{array}$

17. यदि $A, 3 \times 3$ कोटी का वर्ग आव्यूह है तो $| a d j A |$ का मान है।

(a) $| A |$

(b) ${| A |}^{2}$

(d) $3 | A |$

उत्तर: (b) ${| A |}^{2}$

18. यदि $A$ कोटी दो का व्युत्क्रमिय आव्यूह है तो $d e t (A^{- 1})$ बराबर:

(a) $d e t (A)$

(b) $\frac{1}{d e t (A)}$

(d) $0$

उत्तर: (b) $\frac{1}{d e t (A)}$

प्रश्नावली 4.6

निम्नलिखित प्रश्नों 1 से 6 तक दी गई समीकरण निकायों का संगत अथवा असंगत के वर्गीकरण कीजिए:

1. $x + 2 y = 2, 2 x + 3 y = 3$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{l} 1 & 2 \\ 2 & 3 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{l} 2 \\ 3 \end{array}] \\ | A | = | \begin{array}{l} 1 & 2 \\ 2 & 3 \end{array} | = 1 \times 3 - 2 \times 2 = - 1 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है। तो यह समीकरण निकाय संगत है।

2. $2 x - y = 5, x + y = 4$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 2 & - 1 \\ 1 & 1 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{l} 5 \\ 4 \end{array}] \\ | A | = | \begin{array}{c} 2 & - 1 \\ 1 & 1 \end{array} | = 2 \times 1 - (- 1) \times 1 = 3 \neq 0 \end{array}$

3. $x + 3 y = 5, 2 x + 6 y = 8$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{l} 1 & 3 \\ 2 & 6 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{l} 5 \\ 8 \end{array}] \\ | A | = | \begin{array}{l} 1 & 3 \\ 2 & 6 \end{array} | = 1 \times 6 - 3 \times 2 = 0 \end{array}$

$A d j A = [\begin{matrix} 6 & - 2 \\ - 3 & 1 \end{matrix}]$

$(A d j A) B = [\begin{matrix} 6 & - 2 \\ - 3 & 1 \end{matrix}] [\begin{array}{l} 5 \\ 8 \end{array}] = [\begin{array}{l} 14 \\ - 7 \end{array}] \neq 0$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व नहीं है। तो यह समीकरण निकाय संगत है।

4. $x + y + z = 1, 2 x + 3 y + 2 z = 2, a x + a y + 2 a z = 4$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{l} 1 \\ 2 \\ 4 \end{array}] \\ | A | = | \begin{array}{c} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{array} | = 1 (3 \times 2 a - 2 \times a) - 1 (2 \times 2 a - 2 \times a) + (2 \times a - 3 \times a) \\ = a \neq 0 \end{array}$

5. $3 x - y - 2 z = 2, 2 y - z = - 1, 3 x - 5 y = 3$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{c} 2 \\ - 1 \\ 3 \end{array}] \\ | A | = | \begin{array}{c} 3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{array} | = - 15 + 3 + 12 = 0 \\ A d j A = [\begin{array}{c} - 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6 \end{array}] \\ (A d j A) B = [\begin{array}{c} - 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6 \end{array}] [\begin{array}{c} 2 \\ - 1 \\ 3 \end{array}] = [\begin{array}{c} - 10 - 10 + 15 \\ - 6 - 6 + 9 \\ - 12 - 12 + 18 \end{array}] \end{array}$

$= [\begin{array}{l} - 5 \\ - 3 \\ - 6 \end{array}] \neq 0$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व नहीं है। तो यह समीकरण निकाय असंगत है।

6. $5 x - y + 4 z = 5, 2 x + 3 y - 5 z = 2, 5 x - 2 y + 6 z = - 1$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{c} 5 \\ 2 \\ - 1 \end{array}] \\ | A | = | \begin{array}{c} 5 & - 1 & 4 \\ 2 & 3 & 5 \\ 5 & - 2 & 6 \end{array} | = 5 (18 + 3) - 1 (12 - 25) + 4 (- 4 - 15) \\ = 5 \times 28 - 13 - 4 \times 5 = 67 \neq 0 \end{array}$

निम्नलिखित प्रश्न 7 से 17 तक प्रत्येक समीकरण निकाय को आव्यूह विधि से हल कीजिए:

7. $5 x + 2 y = 4, 7 x + 3 y = 5$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{l} 5 & 2 \\ 7 & 3 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{l} 4 \\ 5 \end{array}] \\ | A | = | \begin{array}{l} 5 & 2 \\ 7 & 3 \end{array} | = 5 \times 3 - 7 \times 2 = 15 - 14 = 1 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{c} 3 & - 2 \\ - 7 & 5 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = [\begin{array}{c} 3 & - 2 \\ - 7 & 5 \end{array}] \end{array}$

$X = A^{- 1} B = [\begin{matrix} 3 & - 2 \\ - 7 & 5 \end{matrix}] [\begin{array}{l} 4 \\ 5 \end{array}] = [\begin{matrix} (12 - 10) \\ (- 28 + 25) \end{matrix}] = [\begin{matrix} 2 \\ - 3 \end{matrix}]$

$[\begin{array}{l} x \\ y \end{array}] = [\begin{matrix} 2 \\ - 3 \end{matrix}]$

इसलिए इस समीकरण निकाय का हल है:

$x = 2, y = - 3$

8. $2 x - y = - 2, 3 x + 4 y = 3$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 2 & - 1 \\ 3 & 4 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{c} - 2 \\ 3 \end{array}] \\ | A | = | \begin{array}{c} 2 & - 1 \\ 3 & 4 \end{array} | = 2 \times 4 + 1 \times 3 = 8 + 3 = 11 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{c} 4 & 1 \\ - 3 & 2 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{11} [\begin{array}{c} 4 & 1 \\ - 3 & 2 \end{array}] \end{array}$

$\begin{array}{r} X = A^{- 1} B = \frac{1}{11} \\ [\begin{array}{c} 4 & 1 \\ - 3 & 2 \end{array}] [\begin{array}{c} - 2 \\ 3 \end{array}] = [\begin{array}{c} (4 \times - 2 + 1 \times 3) \\ (- 3 \times - 2 + 2 \times 3) \end{array}] = \frac{1}{11} [\begin{array}{c} - 5 \\ 12 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{c} \frac{- 5}{11} \\ \frac{12}{11} \end{array}] \end{array}$

इसलिए इस समीकरण निकाय का हल है:

$x = - \frac{5}{11}, y = \frac{12}{11}$

9. $4 x - 3 y = 3, 3 x - 5 y = 7$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{l} 4 & - 3 \\ 3 & - 5 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{l} 3 \\ 7 \end{array}] \\ | A | = | \begin{array}{l} 4 & - 3 \\ 3 & - 5 \end{array} | = 4 \times - 5 + 3 \times 3 = - 20 + 9 = - 11 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{l} - 5 & 3 \\ - 3 & 4 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = - \frac{1}{11} [\begin{array}{l} - 5 & 3 \\ - 3 & 4 \end{array}] \\ X = A^{- 1} B = \frac{1}{11} \\ [\begin{array}{l} - 5 & 3 \\ - 3 & 4 \end{array}] [\begin{array}{l} 3 \\ 7 \end{array}] = [\begin{array}{l} (- 5 \times 3 + 3 \times 7) \\ (- 3 \times 3 + 4 \times 7) \end{array}] = - \frac{1}{11} [\begin{array}{c} 6 \\ 19 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{l} - \frac{6}{11} \\ - \frac{19}{11} \end{array}] \end{array}$

इसलिए इस समीकरण निकाय का हल है:

$x = \frac{6}{11}, y = - \frac{19}{11}$

10. $5 x + 2 y = 3, 3 x + 2 y = 5$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{l} 5 & 2 \\ 3 & 2 \end{array}], X = [\begin{array}{l} x \\ y \end{array}], B = [\begin{array}{l} 3 \\ 7 \end{array}] \\ | A | = | \begin{array}{l} 5 & 2 \\ 3 & 2 \end{array} | = 5 \times 2 - 3 \times 2 \end{array}$

$= 10 - 6 = 4 \neq 0$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$A d j A = [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}]$

$A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{4} [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}]$

$\begin{array}{r} X = A^{- 1} B = \frac{1}{4} [\begin{array}{c} 2 & - 2 \\ - 3 & 5 \end{array}] [\begin{array}{l} 3 \\ 5 \end{array}] = [\begin{array}{c} (2 \times 3 - 2 \times 5) \\ (- 3 \times 3 + 5 \times 5) \end{array}] = \frac{1}{4} [\begin{array}{c} - 4 \\ 16 \end{array}] \\ [\begin{array}{l} x \\ y \end{array}] = [\begin{array}{c} - 1 \\ 16 \end{array}] \end{array}$

इसलिए इस समीकरण निकाय का हल है:

$x = - 1, y = 4$

11. $2 x + y + z = 1, x - 2 y - z = \frac{3}{2}, 3 y - 5 z = 9$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A = [\begin{array}{c} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{l} 1 \\ \frac{3}{2} \\ 9 \end{array}] \\ | A | = | \begin{array}{c} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{array} | = 2 (10 + 3) - 1 (- 5 - 3) + 0 (- 1 + 3) \\ = 26 + 8 + 0 = 34 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{c} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{34} [\begin{array}{c} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{array}] \end{array}$

$X = A^{- 1} B$

$= \frac{1}{34} [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}] [\begin{matrix} 1 \\ \frac{3}{2} \\ 9 \end{matrix}] = \frac{1}{34} [\begin{matrix} (13 \times 1 + 8 \times \frac{3}{2} + 1 \times 9) \\ (5 \times 1 - 10 \times \frac{3}{2} + 3 \times 9) \\ (3 \times 1 - 6 \times \frac{3}{2} - 5 \times 9) \end{matrix}]$

$= \frac{1}{34} [\begin{matrix} 34 \\ 17 \\ - 51 \end{matrix}]$

$[\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{matrix} 1 \\ \frac{1}{2} \\ - \frac{3}{2} \end{matrix}]$

इसलिए इस समीकरण निकाय का हल है:

$x = 1, y = \frac{1}{2}, Z = - \frac{3}{2}$

12. $x - y + z = 4, 2 x + y - 3 z = 0, x + y + z = 2$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{l} 4 \\ 0 \\ 2 \end{array}] \\ | A | = | \begin{array}{c} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 11 & 1 & 1 \end{array} | = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) \\ = 4 + 5 + 1 = 10 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{c} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{10} [\begin{array}{c} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{array}] \\ X = A^{- 1} B \\ = \frac{1}{10} [\begin{array}{c} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{array}] [\begin{array}{l} 4 \\ 0 \\ 2 \end{array}] = \frac{1}{10} [\begin{array}{c} (4 \times 4 + 2 \times 0 + 2 \times 2) \\ (- 5 \times 4 + 0 \times 0 + 5 \times 2) \\ (1 \times 4 - 2 \times 0 - 3 \times 2) \end{array}] \\ = \frac{1}{10} [\begin{array}{c} 16 + 0 + 4 \\ - 20 + 0 + 10 \\ 4 + 0 + 6 \end{array}] = \frac{1}{10} [\begin{array}{c} 20 \\ - 10 \\ 10 \end{array}] \end{array}$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$

इसलिए इस समीकरण निकाय का हल है:

$x = 2, y = - 1, Z = 1$

13. $2 x + 3 y + 3 z = 5, x - 2 y + z = - 4, 3 x - y - 2 z = 3$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 2 & 3 & 3 \\ 1 & - 2 & 1 \\ 3 & - 1 & - 2 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{c} 5 \\ - 4 \\ 3 \end{array}] \\ | A | = | \begin{array}{c} 2 & 3 & 3 \\ 1 & - 2 & 1 \\ 3 & - 1 & - 2 \end{array} | = 2 (4 + 1) - 3 (- 2 - 3) + 3 (- 1 + 6) = 10 + 15 + 15 = 40 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{c} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{40} [\begin{array}{c} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{array}] \\ X = A^{- 1} B = \frac{1}{40} [\begin{array}{c} 5 & 3 & 9 \\ 5 & - 13 & 1 \\ 5 & 11 & - 7 \end{array}] [\begin{array}{c} 5 \\ - 4 \\ 3 \end{array}] = \frac{1}{40} [\begin{array}{c} (5 \times 5 + 3 \times - 4 + 9 \times 3) \\ (5 \times 5 - 13 \times - 4 + 1 \times 3) \\ (5 \times 5 + 11 \times - 4 - 7 \times 3) \end{array}] \end{array}$

$\begin{array}{r} = \frac{1}{40} [\begin{array}{c} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{array}] = \frac{1}{40} [\begin{array}{c} 40 \\ 80 \\ - 40 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{c} 1 \\ 2 \\ - 1 \end{array}] \end{array}$

इसलिए इस समीकरण निकाय का हल है:

$x = 1, y = 2, Z = - 1$

14. $x - y + 2 z = 7, 3 x + 4 y - 5 z = - 5, 2 x - y + 3 z = 12$

उत्तर: दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है:

$\begin{array}{r} A X = B \\ A = [\begin{array}{c} 1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & - 3 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{c} 7 \\ - 5 \\ 12 \end{array}] \\ | A | = | \begin{array}{c} 1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & - 3 \end{array} | \\ = 1 (12 - 5) + 1 (9 + 10) + 2 (- 3 - 8) \\ = 7 + 19 - 22 = 4 \neq 0 \end{array}$

इसलिए A एक व्युत्क्रमनीय आव्यूह है तथा इसके व्युत्क्रम का अस्तित्व है।

$\begin{array}{r} A d j A = [\begin{array}{c} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{4} [\begin{array}{c} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{array}] \\ X = A^{- 1} B \\ = \frac{1}{4} [\begin{array}{c} 7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7 \end{array}] [\begin{array}{c} 7 \\ - 5 \\ 12 \end{array}] = \frac{1}{4} [\begin{array}{c} (7 \times 7 + 1 \times - 5 - 3 \times 12) \\ (- 19 \times 7 - 1 \times - 5 + 11 \times 12) \\ (- 11 \times 7 - 1 \times - 5 + 7 \times 12) \end{array}] \\ = \frac{1}{4} [\begin{array}{c} 49 - 5 - 36 \\ - 133 + 5 + 132 \\ - 77 + 5 + 84 \end{array}] = \frac{1}{4} [\begin{array}{c} 8 \\ 4 \\ 12 \end{array}] \\ [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{l} 2 \\ 1 \\ 3 \end{array}] \end{array}$

इसलिए इस समीकरण निकाय का हल है:

$x = 2, y = 1, Z = 3$

15. यदि $A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}]$ है तो $A^{- 1}$ ज्ञात कीजिए, $A^{- 1}$ का प्रयोग करके निम्नलिखित समीकरण निकाय को हल कीजिए:

$2 x - 3 y + 5 z = 11, 3 x + 2 y - 4 z = - 5, x + y - 2 z = - 3$

उत्तर: $A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}]$

$\begin{array}{r} | A | = | \begin{array}{c} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{array} | = 2 (- 4 + 4) + 3 (- 6 + 4) + 5 (3 - 2) \\ = 0 - 6 + 5 = - 1 \\ A d j A = [\begin{array}{c} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}] \\ A^{- 1} = - [\begin{array}{l} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{array}] = [\begin{matrix} - 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{matrix}] \\ ∵ X = A^{- 1} B \\ ∴ B = [\begin{array}{c} 11 \\ - 5 \\ - 3 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] \\ 2 x - 3 y + 5 z = 11, 3 x + 2 y - 4 z = - 5, x + y - 2 z = - 3 \end{array}$

$\begin{array}{r} X = [\begin{array}{l} - 0 & 1 & - 2 \\ - 2 & 9 & - 23 \\ - 1 & 5 & - 13 \end{array}] [\begin{array}{l} 11 \\ - 5 \\ - 3 \end{array}] \\ = [\begin{array}{l} (- 0 \times 11 - 1 \times 5 + 2 \times 3) \\ (- 2 \times 11 - 9 \times 5 + 23 \times 3) \\ (- 1 \times 11 - 5 \times 5 + 13 \times 3) \end{array}] = [\begin{array}{c} 0 - 5 + 6 \\ - 22 - 45 + 69 \\ - 11 - 25 + 39 \end{array}] \end{array}$

$= [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$

$[\begin{array}{l} y \\ z \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}]$

इसलिए, इस समीकरण निकाय का हल है

$x = 1, y = 2, Z = 3$

16. $4 k g$ प्याज, $3 k g$ गेहू और $2 k g$ चावल का मूल्य $R s . 60$ है। $2 k g$ प्याज, $4 k g$ गेहू और $6 k g$ चावल का मूल्य $R s . 90$ है। $6 k g$ प्याज, $2 k g$ गेहू और $3 k g$ चावल का मूल्य $R s . 70$ है। आव्यूह विधि द्वारा प्रत्येक का मूल्य प्रति $k g$ ज्ञात कीजिए।

उत्तर: मान लेते है की प्याज का मूल्य प्रति $k g R s . x$ , गेहू का मूल्य प्रति $k g R s . y$ , तथा चवक का मूल्य प्रति $k g R s . z$ है।

इसलिए हम फी गई जानकारी को समीकरणों के रूप मे कुछ इस प्रकार रख सकते है:

$4 x + 3 y + 2 z = 60$

$\begin{array}{r} 2 x + 4 y + 6 z = 90 \\ 6 x + 2 y + 3 y = 70 \end{array}$

दिए गए समीकरण निकायों को इस प्रकार से लिख सकते है।

$\begin{array}{r} A X = B \\ A = [\begin{array}{l} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{l} 60 \\ 90 \\ 70 \end{array}] \\ | A | = | \begin{array}{l} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{array} | = 4 (12 - 12) - 3 (6 - 36) + 2 (4 - 24) = 0 + 90 - 40 = 50 \neq 0 \\ A d j A = [\begin{array}{c} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array}] \\ A^{- 1} = \frac{A d j A}{| A |} = \frac{1}{50} [\begin{array}{c} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array}] \\ X = A^{- 1} B = \frac{1}{50} [\begin{array}{c} 0 & - 5 & 10 \\ 30 & 0 & - 20 \\ - 20 & 10 & 10 \end{array}] [\begin{array}{l} 60 \\ 90 \\ 70 \end{array}] \end{array}$

$= \frac{1}{50} [\begin{matrix} (0 \times 60 - 5 \times 90 + 10 \times 70) \\ (30 \times 60 + 0 \times 90 - 20 \times 70) \\ (- 20 \times 60 + 10 \times 90 + 10 \times 70) \end{matrix}] = \frac{1}{50} [\begin{matrix} 0 - 450 + 700 \\ 1800 + 0 - 1400 \\ - 1200 + 900 + 700 \end{matrix}]$

$\begin{array}{r} = \frac{1}{50} [\begin{array}{l} 250 \\ 400 \\ 400 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 5 \\ 8 \\ 8 \end{array}] \end{array}$

इसलिए इस समीकरण निकाय का हल है:

$x = 5, y = 8, Z = 8$

प्याज का मूल्य प्रति $k g R s . 5$ , गेहू का मूल्य प्रति $k g R s . 8$ , तथा चावल का मूल्य प्रति $k g R s . 8$ है।

प्रश्नावली A4

1. सिद्ध कीजिए की सारणिक $| \begin{matrix} x & s i n θ & c o s θ \\ - s i n θ & - x & 1 \\ c o s θ & 1 & x \end{matrix} | θ$ से स्वतंत्र है।

उत्तर: $Δ = | \begin{matrix} x & s i n θ & c o s θ \\ - s i n θ & - x & 1 \\ c o s θ & 1 & x \end{matrix} |$

$\begin{array}{r} = x - (x^{2} - 1) - s i n θ (- x s i n θ - c o s θ) + c o s θ (- s i n θ + x c o s θ) \\ = - x^{3} - x + x s i n^{2} θ + s i n θ c o s θ - c o s θ s i n θ + x c o s^{2} θ \\ = - x^{3} - x + x (s i n^{2} θ + c o s^{2} θ) \\ = - x^{3} - x + x \\ = - x^{2} \end{array}$

जो $θ$ से स्वतंत्र है।

2. सारणिक का प्रसरण किए बिना सिद्ध कीजिए की $| \begin{array}{l} a & a^{2} & b c \\ b & b^{2} & b c \\ c & c^{2} & a b \end{array} | = | \begin{array}{l} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{array} |$

उत्तर: $| \begin{array}{l} a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b \end{array} | = \frac{1}{a b c} | \begin{array}{l} a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c \end{array} | [R_{1} \to a R_{1}, R_{2} \to b R_{2}, R_{3} \to c R_{3}]$

$= \frac{a b c}{a b c} | \begin{array}{l} a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1 \end{array} |$ [ $C_{3}$ से $a b c$ आम लेना]

$\begin{array}{r} = (- 1)^{2} | \begin{array}{l} 1 & a^{3} & a^{2} \\ 1 & b^{3} & b^{2} \\ 1 & c^{3} & c^{2} \end{array} | [C_{1} \leftrightarrow C_{3}] \\ = (- 1)^{2} | \begin{array}{l} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{array} | [C_{2} \leftrightarrow C_{3}] \end{array}$

$= | \begin{array}{l} 1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3} \end{array} |$

3. $| \begin{matrix} c o s α c o s β & c o s α s i n β & - s i n α \\ - s i n β & c o s β & 0 \\ s i n α c o s β & s i n α s i n β & c o s α \end{matrix} |$ का मान ज्ञात कीजिए।

उत्तर: $= - s i n α (- s i n α s i n^{2} β - s i n α c o s^{2} β) - 0 (c o s α c o s β s i n α s i n β - c o s α s i n β s i n α c o s β)$

$+ c o s α (c o s α c o s^{2} β + c o s α c o s^{2} β)$

$\begin{array}{r} = s i n^{2} α (s i n^{2} β + c o s^{2} β) + c o s^{2} α (c o s^{2} β + s i n^{2} β) \\ = s i n^{2} α + c o s^{2} α \\ = 1 \end{array}$

4. यदि $a, b, c$ वास्तविक संखयाए हो और सारणिक $Δ = | \begin{array}{l} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{array} | = 0$ हो तो दर्शाइए की या तो $a + b + c = 0$ या $a = b = c$ है।

उत्तर: $| \begin{array}{l} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{array} | = 0$

$= | \begin{matrix} 2 (a + b + c) & 2 (a + b + c) & 2 (a + b + c) \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0 [R_{1} \to R_{1} + R_{2} + R_{3}]$

$= 2 (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0$

[ $R_{1}$ के सामान्य रूप से $2 (a + b + c)$ ]

$= 2 (a + b + c) | \begin{matrix} 1 & 0 & 0 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{matrix} | = 0 [C_{2} \to C_{2} - C 1, C_{3} \to C_{3} - C_{1}]$

$\begin{array}{r} ∴ 2 (a + b + c) [(b - c) (c - b) - (b - a) (c - a)] = 0 \\ 2 (a + b + c) [b c - b^{2} - c^{2} + b c - (b c - a b - a c + a^{2})] = 0 \end{array}$

$\begin{array}{r} 2 (a + b + c) [b c - b^{2} - c^{2} + b c - b c + a b + a c - a^{2}] = 0 \\ - (a + b + c) [2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a] = 0 \end{array}$

$\begin{array}{r} - (a + b + c) [(a^{2} + b^{2} + 2 a b) + (b^{2} + c^{2} - 2 b c) + (c^{2} + a^{2} - 2 c a)] = 0 \\ - (a + b + c) [{(a - b)}^{2} + (b - c)^{2} + (c - a)^{2}] = 0 \end{array}$

$a + b + c = 0; (a - b)^{2} = 0, (b - c)^{2} = 0, (c - a)^{2} = 0$

$\begin{array}{r} a + b + c = 0; (a - b) = 0, (b - c) = 0, (c - a) = 0 \\ a + b + c = 0; a = b, b = c, c = a \\ a + b + c = 0; a = b = c \end{array}$

5. यदि $a \neq 0$ हो तो समीकरण $| \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0$ को हल कीजिए।

उत्तर: $| \begin{matrix} x + a & x & x \\ x & x + a & x \\ x & x & x + a \end{matrix} | = 0$

$= | \begin{matrix} 3 x + a & 3 x + a & 3 x + a \\ x & x + 2 & x \\ x & x & x + a \end{matrix} | = 0 [R_{1} \to R_{1} + R_{2} + R_{3}]$

$= (3 x + a) | \begin{matrix} 1 & 1 & 1 \\ x & x + 2 & x \\ x & x & x + a \end{matrix} | = 0$ [ $R_{1}$ से सामान्य रूप मे $(3 x + a)$ ले]

$\begin{array}{r} = (3 x + a) | \begin{array}{l} 1 & 0 & 0 \\ x & a & 0 \\ x & 0 & a \end{array} | = 0 [C_{2} \to C_{2} - C_{1}, C_{3} \to C_{3} - C_{1}] \\ = (3 x + a) [a^{2} - 0] \\ = 0 \end{array}$

$\begin{array}{r} = a^{2} (3 x + a) = 0 \\ \Rightarrow (3 x + a) = 0 [∵ a \neq 0] \\ \Rightarrow x = - \frac{a}{3} \end{array}$

6. सिद्ध कीजिए की $| \begin{matrix} a^{2} & b c & a c + c^{2} \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix} | = 4 a^{2} b^{2} c^{2}$

उत्तर: $| \begin{matrix} a^{2} & b c & a c + c^{2} \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix} | = | \begin{matrix} 2 a^{2} & 0 & 2 a c \\ a^{2} + a b & b^{2} & a c \\ a b & b^{2} + b c & c^{2} \end{matrix} |$

$= a b c | \begin{matrix} 2 a & 0 & 2 a \\ a + b & b & a \\ b & b + c & c \end{matrix} | [R_{1} \to R_{2} + R_{2} - R_{3}]$

$\begin{array}{r} = a b c | \begin{array}{c} 2 a & 0 & 0 \\ a + b & b & - b \\ b & b + c & c - b \end{array} | \\ = (a b c) 2 a [b (c - a) - b (b + c)] \\ = 2 a^{2} b c [b c - b^{2} + b^{2} + b c] \\ = 2 a^{2} b c [2 b c] \\ = 4 a^{2} b^{2} c^{2} \end{array}$

7. $A^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ और $B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$ हो तो ${(A B)}^{- 1}$ का मान ज्ञात कीजिए।

उत्तर: $B = [\begin{matrix} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{matrix}]$

$\begin{matrix} B_{11} = 3 & B_{12} = 1 & B_{13} = 2 \\ B_{21} = 2 & B_{22} = 1 & B_{23} = 2 \\ B_{31} = 6 & B_{32} = 2 & B_{33} = 5 \end{matrix}$

$\begin{array}{r} B^{- 1} = \frac{1}{| B |} a d j B \\ = \frac{1}{1} [\begin{array}{l} B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{23} \\ B_{31} & B_{32} & B_{33} \end{array}] = [\begin{array}{c} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}] \\ ∴ {(A B)}^{- 1} = B^{- 1} A^{- 1} \\ {(A B)}^{- 1} = B^{- 1} A^{- 1} = [\begin{array}{l} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}] = [\begin{array}{c} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{array}] \\ = [\begin{array}{c} 9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & - 2 + 12 - 10 & 2 - 10 + 10 \end{array}] \\ = [\begin{array}{c} 9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2 \end{array}] \end{array}$

8. मान लीजिए $[\begin{matrix} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$ हो तो सत्यापित कीजिए की

(i) ${[a d j A]}^{- 1} = a d j (A^{- 1})$

उत्तर: $A = [\begin{matrix} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{matrix}]$

$| A | = 1 (15 - 1) + 2 (- 10 - 1) + 1 (- 2 - 3) = - 13 \neq 0 \Rightarrow A^{- 1}$

$\begin{matrix} A_{11} = 14 & A_{12} = 11 & A_{13} = - 5 \\ A_{21} = 11 & A_{22} = 4 & A_{23} = - 3 \\ A_{31} = - 5 & A_{32} = - 3 & A_{33} = - 1 \end{matrix}$

$A^{- 1} = \frac{1}{| A |} a d j A = \frac{1}{| A |} [\begin{array}{l} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}] = \frac{1}{- 13} [\begin{matrix} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{matrix}]$ ………………..(1)

$\begin{array}{r} B = a d j A \\ B = [\begin{array}{c} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{array}] \\ | B | = 14 (- 4 - 9) - 11 (- 11 - 15) - 5 (- 33 + 20) = - 182 + 286 + 65 = 169 \neq 0 \Rightarrow B^{- 1} \end{array}$

$\begin{matrix} B_{11} = - 13 & B_{12} = 26 & B_{13} = - 13 \\ B_{21} = 26 & B_{22} = - 39 & B_{23} = - 13 \\ B_{31} = - 13 & B_{32} = - 13 & B_{33} = - 65 \end{matrix}$

$B^{- 1} = \frac{1}{| B |} a d j B = \frac{1}{| B |} [\begin{array}{l} B_{11} & B_{12} & B_{13} \\ B_{21} & B_{22} & B_{23} \\ B_{31} & B_{32} & B_{33} \end{array}] = \frac{1}{169} [\begin{matrix} - 13 & 26 & - 13 \\ 26 & - 39 & - 13 \\ - 13 & - 13 & - 65 \end{matrix}] = \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}]$

$a d j A^{- 1} = \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}]$ …………………..(2)

$\begin{array}{r} C = A^{- 1} \\ C = \frac{1}{- 13} [\begin{array}{c} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{array}] \end{array}$

$\begin{matrix} C_{11} = 3 & C_{12} = 1 & C_{13} = 2 \\ C_{21} = 2 & C_{22} = 1 & C_{23} = 2 \\ C_{31} = 6 & C_{32} = 2 & C_{33} = 5 \end{matrix}$

$a d j C = \frac{1}{1} [\begin{array}{l} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{array}] = [\begin{matrix} - \frac{1}{13} & \frac{2}{13} & - \frac{1}{13} \\ \frac{2}{13} & - \frac{3}{13} & - \frac{1}{13} \\ - \frac{1}{13} & - \frac{1}{13} & - \frac{5}{13} \end{matrix}] = \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}]$

$a d j C = a d j (A^{- 1}) = \frac{1}{13} [\begin{matrix} - 1 & 2 & - 1 \\ 2 & - 3 & - 1 \\ - 1 & - 1 & - 5 \end{matrix}]$ ………………..(3)

समीकरण (2) और (3) से हमारे पास है ${(a d j A)}^{- 1} = a d j (A^{- 1})$

(ii) ${(A^{- 1})}^{- 1} = A$

उत्तर: समीकरण (1) से हमारे पास

$\begin{array}{r} A^{- 1} = \frac{1}{- 13} [\begin{array}{c} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{array}] \\ D = A^{- 1} = \frac{1}{- 13} [\begin{array}{c} 14 & 11 & - 5 \\ 11 & 4 & - 3 \\ - 5 & - 3 & - 1 \end{array}] = [\begin{array}{c} \frac{14}{13} & \frac{11}{13} & - \frac{5}{13} \\ \frac{11}{13} & \frac{4}{13} & - \frac{3}{13} \\ - \frac{5}{13} & - \frac{3}{13} & - \frac{1}{13} \end{array}] \\ | D | = - {(\frac{1}{13})}^{3} [14 (- 4 - 9) - 11 (- 11 - 15) - 5 (- 33 + 20)] \\ = - {(\frac{1}{13})}^{3} (- 182 + 286 + 65) = - {(\frac{1}{13})}^{3} 169 = - (\frac{1}{13}) \neq 0 \Rightarrow D^{- 1} \end{array}$

$\begin{array}{l} D_{11} = - (\frac{1}{13}) & D_{12} = - (\frac{1}{13}) & D_{13} = - (\frac{1}{13}) \\ D_{21} = - (\frac{1}{13}) & D_{22} = - (\frac{1}{13}) & D_{23} = - (\frac{1}{13}) \\ D_{31} = - (\frac{1}{13}) & D_{23} = - (\frac{1}{13}) & D_{33} = - (\frac{1}{13}) \end{array}$

$\begin{array}{r} D^{- 1} = \frac{1}{| D |} a d j D = \frac{1}{| D |} [\begin{array}{l} D_{11} & D_{12} & D_{13} \\ D_{21} & D_{22} & D_{23} \\ D_{31} & D_{32} & D_{33} \end{array}] \\ = \frac{1}{- 1} [\begin{array}{c} - \frac{1}{13} & \frac{2}{13} & - \frac{1}{13} \\ \frac{2}{13} & - \frac{3}{13} & - \frac{1}{13} \\ - \frac{1}{13} & - \frac{1}{13} & - \frac{5}{13} \end{array}] = [\begin{array}{c} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{array}] \\ D^{- 1} = {(A^{- 1})}^{- 1} = | \begin{array}{c} 1 & - 2 & 1 \\ - 2 & 3 & 1 \\ 1 & 1 & 5 \end{array} | = A \end{array}$

9. $| \begin{matrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{matrix} |$ का मान ज्ञात कीजिए।

उत्तर: $= | \begin{matrix} 2 (x + y) & y & x + y \\ 2 (x + y) & x + y & x \\ 2 (x + y) & x & y \end{matrix} | [C_{1} \to C_{1} + C_{2} + C_{3}]$

$= 2 (x + y) | \begin{matrix} 1 & y & x + y \\ 1 & x + y & x \\ 1 & x & y \end{matrix} |$ [ $C_{1}$ से सामान्य के रूप मे $2 (x + y)$ लेना]

$\begin{array}{r} = 2 (x + y) | \begin{array}{c} 0 & - x & y \\ 0 & y & x - y \\ 1 & x & y \end{array} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}] \\ = 2 (x + y) [(- x) (x - y) - y \times y] \\ = 2 (x + y) (- x^{2} + x y - y^{2}) \\ = - 2 (x + y) (x^{2} - x y + y^{2}) \\ = - 2 (x^{2} + y^{2}) \end{array}$

10. $| \begin{matrix} 1 & x & x + y \\ 1 & x + y & x \\ 1 & x & y \end{matrix} |$ का मान ज्ञात कीजिए।

उत्तर: $= | \begin{matrix} 0 & - y & 0 \\ 0 & y & - x \\ 1 & x & x + y \end{matrix} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$\begin{array}{r} = [(- y) (- x) - y \times 0] \\ = x y \end{array}$

सारणिकों के गुणधर्मों का प्रयोग करके निम्नलिखित 11 से 15 तक प्रश्नों को सिद्ध कीजिए:

11. $| \begin{array}{l} α & α^{2} & β + γ \\ β & β^{2} & γ + α \\ γ & γ^{2} & α + β \end{array} | = (β - γ) (γ - α) (α - β) (α + β + γ)$

उत्तर: LHS = $| \begin{array}{l} α & α^{2} & β + γ \\ β & β^{2} & γ + α \\ γ & γ^{2} & α + β \end{array} |$

$= | \begin{array}{l} α & α^{2} & α + β + γ \\ β & β^{2} & α + β + γ \\ γ & γ^{2} & α + β + γ \end{array} | [C_{3} \to C_{3} + C_{2}]$

$= (α + β + γ) | \begin{array}{l} α & α^{2} & 1 \\ β & β^{2} & 1 \\ γ & γ^{2} & 1 \end{array} |$ [ $C_{3}$ से सामान्य के रूप मे $(α + β + γ)$ लेना]

$= (α + β + γ) | \begin{matrix} α - β & α^{2} - β^{2} & 0 \\ β - γ & β^{2} - γ^{2} & 0 \\ γ & γ^{2} & 1 \end{matrix} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$= (α + β + γ) (α - β) (β - γ) | \begin{matrix} 1 & α + β & 0 \\ 1 & β + γ & 0 \\ γ & γ^{2} & 1 \end{matrix} |$ [ $R_{1}$ से सामान्य के रूप मे $(α - β)$ लेना और $R_{2}$ से सामान्य रूप से $(β - γ)$ लेना]

$\begin{array}{r} = (α + β + γ) (α - β) (β - γ) [(β + γ) - (α + β)] \\ = (α + β + γ) (α - β) (β - γ) (γ - α) \end{array}$

= RHS

12. $| \begin{array}{l} x & x^{2} & 1 + p x^{2} \\ y & y^{2} & 1 + p y^{2} \\ z & z^{2} & 1 + p z^{2} \end{array} | = (1 + p x y z) (x - y) (y - z) (z - x)$

उत्तर: LHS = $| \begin{array}{l} x & x^{2} & 1 + p x^{2} \\ y & y^{2} & 1 + p y^{2} \\ z & z^{2} & 1 + p z^{2} \end{array} |$

$= | \begin{array}{l} x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1 \end{array} | + | \begin{array}{l} x & x^{2} & p x^{2} \\ y & y^{2} & p y^{2} \\ z & z^{2} & p z^{2} \end{array} |$

$= (- 1)^{1} | \begin{array}{l} 1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z \end{array} | + p | \begin{array}{l} x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3} \end{array} |$

[ $C_{3}$ से सामान्य के रूप मे $p$ लेना]

$= (- 1)^{2} | \begin{array}{l} 1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z \end{array} | + p x y z | \begin{array}{l} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array} |$ [ $R_{1}, R_{2}, R_{3}$ से सामान्य के रूप मे $x, y, z$ लेना]

$= (1 + p x y z) | \begin{array}{l} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array} |$

[ $| \begin{array}{l} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array} |$ सामान्य के रूप मे लेना]

$= (1 + p x y z) | \begin{matrix} 0 & x - y & x^{2} - y^{2} \\ 0 & y - z & y^{2} - z^{2} \\ 1 & z & z^{2} \end{matrix} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$= (1 + p x y z) (x - y) (y - z) | \begin{matrix} 0 & 1 & x + y \\ 0 & 1 & y + z \\ 1 & z & z^{2} \end{matrix} |$ [ $R_{1}$ से सामान्य के रूप मे $(x - y)$ लेना और $R_{2}$ से सामान्य के रूप मे $y - z$ लेना]

$\begin{array}{r} = (1 + p x y z) (x - y) (y - z) [(y + z) - (x + y)] \\ = (1 + p x y z) (x - y) (y - z) (z - x) \end{array}$

= RHS

13. $| \begin{matrix} 3 a & - a + b & - a + c \\ - b + a & 3 b & - b + c \\ - c + a & - c + b & 3 c \end{matrix} | = 3 (a + b + c) (a b + b c + c a)$

उत्तर: LHS = $| \begin{matrix} 3 a & - a + b & - a + c \\ - b + a & 3 b & - b + c \\ - c + a & - c + b & 3 c \end{matrix} |$

$= | \begin{matrix} a + b + c & - a + b & - a + c \\ a + b + c & 3 b & - b + c \\ a + b + c & - c + b & 3 c \end{matrix} | [C_{1} \to C_{1} + C_{2} + C_{3}]$

$= (a + b + c) | \begin{matrix} 1 & - a + b & - a + c \\ 1 & 3 b & - b + c \\ 1 & - c + b & 3 c \end{matrix} |$ [ $C_{1}$ से सामान्य के रूप मे $(a + b + c)$ लेना]

$= (a + b + c) | \begin{matrix} 0 & - a - 2 b & - a + c \\ 0 & 2 b + c & - b - 2 c \\ 1 & - c + b & 3 c \end{matrix} | [R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$\begin{array}{r} = (a + b + c) [(- a - 2 b) (- b - 2 c) - (2 b + c) (- a + b)] \\ = (a + b + c) (a b + 2 a c + 2 b^{2} + 4 b c - (- 2 a b + 2 b^{2} = - a c + b c)) \\ = (a + b + c) (3 a b + 3 b c + 3 c a) \\ = 3 (a + b + c) (a b + b c + c a) \end{array}$

= RHS

14. $| \begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} | = 1$

उत्तर: LHS = $| \begin{matrix} 1 & 1 + p & 1 + p + q \\ 2 & 3 + 2 p & 4 + 3 p + 2 q \\ 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} |$

$\begin{array}{r} = | \begin{array}{c} 1 & 1 + p & 1 + p + q \\ 0 & 1 & 2 + p \\ 0 & 3 & 7 + 3 p \end{array} | [R_{2} \to R_{2} - 2 R_{1}, R_{3} \to R_{3} - 3 R_{1}] \\ = 1 (1 (7 + 3 p) - (3) (2 + p)) \\ = 7 + 3 p - 6 - 3 p \\ = 1 \end{array}$

= RHS

15. $| \begin{matrix} s i n α & c o s α & c o s (α + δ) \\ s i n β & c o s β & c o s (β + δ) \\ s i n γ & c o s γ & c o s (γ + δ) \end{matrix} | = 0$

उत्तर: LHS = $| \begin{matrix} s i n α & c o s α & c o s (α + δ) \\ s i n β & c o s β & c o s (β + δ) \\ s i n γ & c o s γ & c o s (γ + δ) \end{matrix} |$

$\begin{array}{r} = | \begin{array}{l} s i n α & c o s α c o s δ - s i n α s i n δ & c o s (α + δ) \\ s i n β & c o s β c o s δ - s i n β s i n δ & c o s (β + δ) \\ s i n γ & c o s γ c o s δ - s i n γ s i n δ & c o s (γ + δ) \end{array} | [C_{2} \to c o s δ C_{2} - s i n δ C_{1}] \\ = | \begin{array}{l} s i n α & c o s (α + δ) & c o s (α + δ) \\ s i n β & c o s (β + δ) & c o s (β + δ) \\ s i n γ & c o s (γ + δ) & c o s (γ + δ) \end{array} | [∵ C_{2} = C_{3}] \\ = 0 \end{array}$

= RHS

16. निम्नलिखित समीकरण निकाय को हल कीजिए:

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$

उत्तर: समीकरणों की इस प्रणाली को $A X = B$ के रूप मे लिखा जा सकता है:

$\begin{array}{r} A = [\begin{array}{c} 2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 7 \end{array}], X = [\begin{array}{l} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{array}], B = [\begin{array}{l} 4 \\ 1 \\ 2 \end{array}] \\ | A | = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) \\ = 150 + 330 + 720 = 1200 \neq 0 \Rightarrow A^{- 1} \end{array}$

$\begin{matrix} A_{11} = 75 & A_{12} = 110 & A_{13} = 72 \\ A_{21} = 150 & A_{22} = - 100 & A_{23} = 0 \\ A_{31} = 75 & A_{32} = 30 & A_{33} = - 24 \end{matrix}$

$A^{- 1} = \frac{1}{| A |} a d j A = \frac{1}{| A |} [\begin{array}{l} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}] = \frac{1}{1200} [\begin{matrix} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{matrix}]$

$\begin{array}{r} X = A^{- 1} B \\ [\begin{array}{l} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{array}] = \frac{1}{1200} [\begin{array}{c} 75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24 \end{array}] [\begin{array}{l} 4 \\ 1 \\ 2 \end{array}] \end{array}$

$\begin{array}{r} [\begin{array}{l} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{array}] = \frac{1}{1200} [\begin{array}{c} 300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48 \end{array}] \\ [\begin{array}{l} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{array}] = \frac{1}{1200} [\begin{array}{l} 600 \\ 400 \\ 240 \end{array}] = [\begin{array}{c} \frac{1}{2} \\ 1 \\ 3 \\ 1 \\ 5 \end{array}] \end{array}$

$\begin{array}{r} \Rightarrow \frac{1}{x} = \frac{1}{2}, \frac{1}{y} = \frac{1}{3}, \frac{1}{z} = \frac{1}{5} \\ \Rightarrow x = 2, y = 3, z = 5 \end{array}$

निम्नलिखित प्रश्नों 17 से 19 मे सही उत्तर का चुनाव कीजिए।

17. यदि $a, b, c$ समातर श्रहई मे हो तो सारणिक $| \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} |$ का मान होगा:

(a) $0$

(b) $1$

(d) $2 x$

उत्तर: $| \begin{matrix} x + 2 & x + 3 & x + 2 a \\ x + 3 & x + 4 & x + 2 b \\ x + 4 & x + 5 & x + 2 c \end{matrix} |$

$= | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ 0 & 0 & 2 (2 b - a - c) \\ x + 4 & x + 5 & x + 2 c \end{matrix} | [R_{2} \to 2 R_{2} - (R_{1} + R_{3})]$

$= | \begin{matrix} x + 2 & x + 3 & x + 2 a \\ 0 & 0 & 0 \\ x + 4 & x + 5 & x + 2 c \end{matrix} |$ [ $a, b, c$ समांतर श्रेणी मे है]

$= 0$

अतः विकल्प (a) सही है।

18. यदि $x, y, z$ शुनएतर वास्तविक संखयाए हो तो आव्यूह $A = | \begin{array}{l} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array} |$ का वयुक्तकर्म है:

(a) $[\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

(b) $x y z [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

(d) $\frac{1}{x y z} [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

उत्तर: $A = [\begin{array}{l} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}]$

$\begin{array}{r} | A | = x (y z - 0) - 0 (0 - 0) + 0 (0 + 0) \\ = x y z \neq 0 \Rightarrow A^{- 1} \end{array}$

$\begin{matrix} A_{11} = y z & A_{12} = 0 & A_{13} = 0 \\ A_{21} = 0 & A_{22} = x z & A_{23} = 0 \\ A_{31} = 0 & A_{32} = 0 & A_{33} = x y \end{matrix}$

$\begin{array}{r} A^{- 1} = \frac{1}{| A |} a d j A = \frac{1}{| A |} [\begin{array}{l} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}] \\ = \frac{1}{x y Z} [\begin{array}{c} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{array}] = [\begin{array}{l} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{array}] \end{array}$

$= [\begin{matrix} x^{- 1} & 0 & 0 \\ 0 & y^{- 1} & 0 \\ 0 & 0 & z^{- 1} \end{matrix}]$

अतः विकल्प (a) सही है।

19. यदि $A = [\begin{matrix} 1 & s i n θ & 1 \\ - s i n θ & 1 & s i n θ \\ - 1 & - s i n θ & 1 \end{matrix}]$ जहा $0 ⩽ θ ⩽ 2 π$ हो तो:

(a) $d e t (A) = 0$

(b) $d e t (A) \in (2, \infty)$

(d) $d e t (A) \in [2, 4]$

उत्तर: $A = [\begin{matrix} 1 & s i n θ & 1 \\ - s i n θ & 1 & s i n θ \\ - 1 & - s i n θ & 1 \end{matrix}]$

$\begin{array}{r} = 1 (1 + s i n^{2} θ) + s i n θ (- s i n θ + s i n θ) + 1 (s i n^{2} θ + 1) \\ = 2 (1 + s i n^{2} θ) \end{array}$

अब यह देखते हुए की $0 ⩽ θ ⩽ 2 π$

$\Rightarrow 0 ⩽ s i n θ ⩽ 1$

$\begin{array}{r} \Rightarrow 0 ⩽ s i n^{2} θ ⩽ 1 \\ \Rightarrow 1 ⩽ 1 + s i n^{2} θ ⩽ 2 \\ \Rightarrow 2 ⩽ 2 (1 + s i n^{2} θ) ⩽ 4 \\ \Rightarrow d e t (A) \in [2, 4] \end{array}$

अतः विकल्प (d) सही है।

NCERT Solutions for Class 12 Maths Chapter 4 Determinants In Hindi

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