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# ${\text{Using elementary transformation, find the inverse of the matrix}} \\ A = \left( {\begin{array}{*{20}{c}} a&b \\ c&{\dfrac{{1 + bc}}{a}} \end{array}} \right) \\ {\text{(a) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\dfrac{{1 + bc}}{a}}&b \\ { - c}&a \end{array}} \right) \\ {\text{(b) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\dfrac{{1 + bc}}{a}}&{ - b} \\ c&a \end{array}} \right) \\ {\text{(c) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\dfrac{{1 + bc}}{a}}&b \\ c&a \end{array}} \right) \\ {\text{(d) None of these}} \\$

Last updated date: 15th Jul 2024
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${\text{We know that}} \\ A = IA \\ {\text{where }}I{\text{ is the inverse matrix such that}} \\ I = \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right) \\ \left( {\begin{array}{*{20}{c}} a&b \\ c&{\dfrac{{1 + bc}}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)A \\$
${{\text{R}}_1} \to \dfrac{{{{\text{R}}_1}}}{a} \\ \left( {\begin{array}{*{20}{c}} 1&{\dfrac{b}{a}} \\ c&{\dfrac{{1 + bc}}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&0 \\ 0&1 \end{array}} \right)A \\$
$\left( {\begin{array}{*{20}{c}} 1&{\dfrac{b}{a}} \\ 0&{\dfrac{{1 + bc}}{a} - \dfrac{{cb}}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&0 \\ {\dfrac{{ - c}}{a}}&1 \end{array}} \right)A \\ \left( {\begin{array}{*{20}{c}} 1&{\dfrac{b}{a}} \\ 0&{\dfrac{1}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&0 \\ {\dfrac{{ - c}}{a}}&1 \end{array}} \right)A \\$
${\text{We have to make }}\dfrac{1}{a} = 1 \\ {{\text{R}}_2} \to a{{\text{R}}_2} \\ \left( {\begin{array}{*{20}{c}} 1&{\dfrac{b}{a}} \\ 0&1 \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} {\dfrac{1}{a}}&0 \\ { - c}&a \end{array}} \right)A \\ {\text{We have to make }}\dfrac{b}{a} = 0 \\ {{\text{R}}_1} \to {{\text{R}}_1} - \dfrac{b}{a}{{\text{R}}_2} \\ \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} {\dfrac{1}{a} - \dfrac{b}{a}( - c)}&{\dfrac{{0 - b}}{a}*a} \\ { - c}&a \end{array}} \right)A \\$
${\text{As we know the identity,}} \\ I = A{A^{ - 1}} \\ {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\dfrac{{1 + bc}}{a}}&{ - b} \\ { - c}&a \end{array}} \right) \\ {\text{So,this is the required answer}}{\text{.}} \\$

${\text{Note:To solve such kind of matrices we should know the identities first}}{\text{.}} \\ {\text{With the use of identities we can solve these questions easily}}{\text{.}} \\$