Answer
Verified
504.3k+ views
\[
{\text{Let }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ be a general determinant }} \\
{\text{As we know that }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ is expanded as,}} \\
\Rightarrow \left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ }} = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|{\text{ }} \\
{\text{This can be reduced as,}} \\
\Rightarrow a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|{\text{ = }}a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right){\text{ }} \\
\Rightarrow {\text{So, }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ }} = {\text{ }}a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right){\text{ (1)}} \\
{\text{Now we know we have to expand }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ }} \\
{\text{So, to expand }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ first we have to compare its elements with the elements of }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ }}. \\
{\text{So on comparing we get }}a = 1,b = - 3,c = 4,d = 3,e = 5,f = - 3,g = 2,h = - 5{\text{ and }}i = 0 \\
{\text{Now putting values of }}a,b,c,d,e,f,g,h{\text{ and }}i{\text{ in equation 3 we get,}} \\
\Rightarrow \left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ }} = {\text{ }}1\left( {5*0 - ( - 5)*( - 3)} \right) + 3\left( {3*0 - 2*( - 3)} \right) + 4\left( {3*( - 5) - 2*5} \right) \\
\Rightarrow {\text{So, }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ = }} - 15 + 18 - 100 = - 97 \\
{\text{Hence }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ }} = - 97{\text{ }} \\
{\text{NOTE: - Whenever you came up to expand a determinant then better way is to expand using cofactors}}{\text{.}} \\
{\text{While expanding calculations should be carefully done}}{\text{.}} \\
\]
{\text{Let }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ be a general determinant }} \\
{\text{As we know that }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ is expanded as,}} \\
\Rightarrow \left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ }} = a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|{\text{ }} \\
{\text{This can be reduced as,}} \\
\Rightarrow a\left| {\begin{array}{*{20}{c}}
e&f \\
h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
d&f \\
g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
d&e \\
g&h
\end{array}} \right|{\text{ = }}a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right){\text{ }} \\
\Rightarrow {\text{So, }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ }} = {\text{ }}a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right){\text{ (1)}} \\
{\text{Now we know we have to expand }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ }} \\
{\text{So, to expand }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ first we have to compare its elements with the elements of }}\left| {\begin{array}{*{20}{c}}
a&b&c \\
d&e&f \\
g&h&i
\end{array}} \right|{\text{ }}. \\
{\text{So on comparing we get }}a = 1,b = - 3,c = 4,d = 3,e = 5,f = - 3,g = 2,h = - 5{\text{ and }}i = 0 \\
{\text{Now putting values of }}a,b,c,d,e,f,g,h{\text{ and }}i{\text{ in equation 3 we get,}} \\
\Rightarrow \left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ }} = {\text{ }}1\left( {5*0 - ( - 5)*( - 3)} \right) + 3\left( {3*0 - 2*( - 3)} \right) + 4\left( {3*( - 5) - 2*5} \right) \\
\Rightarrow {\text{So, }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ = }} - 15 + 18 - 100 = - 97 \\
{\text{Hence }}\left| {\begin{array}{*{20}{c}}
1&{ - 3}&4 \\
3&5&{ - 3} \\
2&{ - 5}&0
\end{array}} \right|{\text{ }} = - 97{\text{ }} \\
{\text{NOTE: - Whenever you came up to expand a determinant then better way is to expand using cofactors}}{\text{.}} \\
{\text{While expanding calculations should be carefully done}}{\text{.}} \\
\]
Recently Updated Pages
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Write the IUPAC name of the given compound class 11 chemistry CBSE
Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
On which part of the tongue most of the taste gets class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Who is the leader of the Lok Sabha A Chief Minister class 11 social science CBSE