\[
{\text{Using elementary transformation, find the inverse of the matrix}} \\
A = \left( {\begin{array}{*{20}{c}}
a&b \\
c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right) \\
{\text{(a) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{\dfrac{{1 + bc}}{a}}&b \\
{ - c}&a
\end{array}} \right) \\
{\text{(b) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{\dfrac{{1 + bc}}{a}}&{ - b} \\
c&a
\end{array}} \right) \\
{\text{(c) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{\dfrac{{1 + bc}}{a}}&b \\
c&a
\end{array}} \right) \\
{\text{(d) None of these}} \\
\]
Last updated date: 21st Mar 2023
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Total views: 312.3k
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Views today: 5.91k
Answer
312.3k+ views
\[
{\text{We know that}} \\
A = IA \\
{\text{where }}I{\text{ is the inverse matrix such that}} \\
I = \left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right) \\
\left( {\begin{array}{*{20}{c}}
a&b \\
c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right)A \\
\]
\[
{{\text{R}}_1} \to \dfrac{{{{\text{R}}_1}}}{a} \\
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
0&1
\end{array}} \right)A \\
\]
\[
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
0&{\dfrac{{1 + bc}}{a} - \dfrac{{cb}}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
{\dfrac{{ - c}}{a}}&1
\end{array}} \right)A \\
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
0&{\dfrac{1}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
{\dfrac{{ - c}}{a}}&1
\end{array}} \right)A \\
\]
\[
{\text{We have to make }}\dfrac{1}{a} = 1 \\
{{\text{R}}_2} \to a{{\text{R}}_2} \\
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
0&1
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
{ - c}&a
\end{array}} \right)A \\
{\text{We have to make }}\dfrac{b}{a} = 0 \\
{{\text{R}}_1} \to {{\text{R}}_1} - \dfrac{b}{a}{{\text{R}}_2} \\
\left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a} - \dfrac{b}{a}( - c)}&{\dfrac{{0 - b}}{a}*a} \\
{ - c}&a
\end{array}} \right)A \\
\]
\[
{\text{As we know the identity,}} \\
I = A{A^{ - 1}} \\
{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{\dfrac{{1 + bc}}{a}}&{ - b} \\
{ - c}&a
\end{array}} \right) \\
{\text{So,this is the required answer}}{\text{.}} \\
\]
\[
{\text{Note:To solve such kind of matrices we should know the identities first}}{\text{.}} \\
{\text{With the use of identities we can solve these questions easily}}{\text{.}} \\
\]
{\text{We know that}} \\
A = IA \\
{\text{where }}I{\text{ is the inverse matrix such that}} \\
I = \left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right) \\
\left( {\begin{array}{*{20}{c}}
a&b \\
c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right)A \\
\]
\[
{{\text{R}}_1} \to \dfrac{{{{\text{R}}_1}}}{a} \\
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
c&{\dfrac{{1 + bc}}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
0&1
\end{array}} \right)A \\
\]
\[
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
0&{\dfrac{{1 + bc}}{a} - \dfrac{{cb}}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
{\dfrac{{ - c}}{a}}&1
\end{array}} \right)A \\
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
0&{\dfrac{1}{a}}
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
{\dfrac{{ - c}}{a}}&1
\end{array}} \right)A \\
\]
\[
{\text{We have to make }}\dfrac{1}{a} = 1 \\
{{\text{R}}_2} \to a{{\text{R}}_2} \\
\left( {\begin{array}{*{20}{c}}
1&{\dfrac{b}{a}} \\
0&1
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a}}&0 \\
{ - c}&a
\end{array}} \right)A \\
{\text{We have to make }}\dfrac{b}{a} = 0 \\
{{\text{R}}_1} \to {{\text{R}}_1} - \dfrac{b}{a}{{\text{R}}_2} \\
\left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}}
{\dfrac{1}{a} - \dfrac{b}{a}( - c)}&{\dfrac{{0 - b}}{a}*a} \\
{ - c}&a
\end{array}} \right)A \\
\]
\[
{\text{As we know the identity,}} \\
I = A{A^{ - 1}} \\
{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}
{\dfrac{{1 + bc}}{a}}&{ - b} \\
{ - c}&a
\end{array}} \right) \\
{\text{So,this is the required answer}}{\text{.}} \\
\]
\[
{\text{Note:To solve such kind of matrices we should know the identities first}}{\text{.}} \\
{\text{With the use of identities we can solve these questions easily}}{\text{.}} \\
\]
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