Question
# $Â Â {\text{Using elementary transformation, find the inverse of the matrix}} \\Â Â A = \left( {\begin{array}{*{20}{c}}Â Â a&b \\Â Â Â c&{\dfrac{{1 + bc}}{a}}Â \end{array}} \right) \\Â Â {\text{(a) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}Â Â {\dfrac{{1 + bc}}{a}}&b \\Â Â Â { - c}&aÂ \end{array}} \right) \\Â Â {\text{(b) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}Â Â {\dfrac{{1 + bc}}{a}}&{ - b} \\Â Â Â c&aÂ \end{array}} \right) \\Â Â {\text{(c) }}{A^{ - 1}} = \left( {\begin{array}{*{20}{c}}Â Â {\dfrac{{1 + bc}}{a}}&b \\Â Â Â c&aÂ \end{array}} \right) \\Â Â {\text{(d) None of these}} \\Â$
$Â Â {\text{We know that}} \\ Â Â A = IA \\ Â Â {\text{where }}I{\text{ is the inverse matrix such that}} \\ Â Â I = \left( {\begin{array}{*{20}{c}} Â Â 1&0 \\ Â Â 0&1 \end{array}} \right) \\ Â Â \left( {\begin{array}{*{20}{c}} Â Â a&b \\ Â Â c&{\dfrac{{1 + bc}}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} Â Â 1&0 \\ Â Â 0&1 \end{array}} \right)A \\$
$Â Â {{\text{R}}_1} \to \dfrac{{{{\text{R}}_1}}}{a} \\ Â Â \left( {\begin{array}{*{20}{c}} Â Â 1&{\dfrac{b}{a}} \\ Â Â c&{\dfrac{{1 + bc}}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} Â Â {\dfrac{1}{a}}&0 \\ Â Â 0&1 \end{array}} \right)A \\ Â$
$Â Â \left( {\begin{array}{*{20}{c}} Â Â 1&{\dfrac{b}{a}} \\ Â Â 0&{\dfrac{{1 + bc}}{a} - \dfrac{{cb}}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} Â Â {\dfrac{1}{a}}&0 \\ Â Â {\dfrac{{ - c}}{a}}&1 \end{array}} \right)A \\ Â Â \left( {\begin{array}{*{20}{c}} Â Â 1&{\dfrac{b}{a}} \\ Â Â 0&{\dfrac{1}{a}} \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} Â Â {\dfrac{1}{a}}&0 \\ Â Â {\dfrac{{ - c}}{a}}&1 \end{array}} \right)A \\$
$Â Â {\text{We have to make }}\dfrac{1}{a} = 1 \\ Â Â {{\text{R}}_2} \to a{{\text{R}}_2} \\ Â Â \left( {\begin{array}{*{20}{c}} Â Â 1&{\dfrac{b}{a}} \\ Â Â 0&1 \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} Â Â {\dfrac{1}{a}}&0 \\ Â Â { - c}&a \end{array}} \right)A \\ Â Â {\text{We have to make }}\dfrac{b}{a} = 0 \\ Â Â {{\text{R}}_1} \to {{\text{R}}_1} - \dfrac{b}{a}{{\text{R}}_2} \\ Â Â \left( {\begin{array}{*{20}{c}} Â Â 1&0 \\ Â Â 0&1 \end{array}} \right){\text{ = }}\left( {\begin{array}{*{20}{c}} Â Â {\dfrac{1}{a} - \dfrac{b}{a}( - c)}&{\dfrac{{0 - b}}{a}*a} \\ Â Â { - c}&a \end{array}} \right)A \\$
$Â Â {\text{As we know the identity,}} \\ Â Â I = A{A^{ - 1}} \\ Â Â {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} Â Â {\dfrac{{1 + bc}}{a}}&{ - b} \\ Â Â { - c}&a \end{array}} \right) \\ Â Â {\text{So,this is the required answer}}{\text{.}} \\$
$Â Â {\text{Note:To solve such kind of matrices we should know the identities first}}{\text{.}} \\ Â Â {\text{With the use of identities we can solve these questions easily}}{\text{.}} \\$