Question

{\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}} \\\

Answer 1

{\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{.}} \\\$$