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

$$\begin{gathered} \text{We know that} \\ A=IA \\ \text{where }I\text{ is the inverse matrix such that} \\ I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ \left(\begin{array}{cc}a& b\\ c& {\displaystyle \frac{1+bc}{a}}\end{array}\right)\text{ = }\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)A \end{gathered}$$
$$\begin{gathered} {\text{R}}_1\to {\displaystyle \frac{{\text{R}}_1}{a}} \\ \left(\begin{array}{cc}1& {\displaystyle \frac{b}{a}}\\ c& {\displaystyle \frac{1+bc}{a}}\end{array}\right)\text{ = }\left(\begin{array}{cc}{\displaystyle \frac{1}{a}}& 0\\ 0& 1\end{array}\right)A \end{gathered}$$
$$\begin{gathered} \left(\begin{array}{cc}1& {\displaystyle \frac{b}{a}}\\ 0& {\displaystyle \frac{1+bc}{a}}-{\displaystyle \frac{cb}{a}}\end{array}\right)\text{ = }\left(\begin{array}{cc}{\displaystyle \frac{1}{a}}& 0\\ {\displaystyle \frac{-c}{a}}& 1\end{array}\right)A \\ \left(\begin{array}{cc}1& {\displaystyle \frac{b}{a}}\\ 0& {\displaystyle \frac{1}{a}}\end{array}\right)\text{ = }\left(\begin{array}{cc}{\displaystyle \frac{1}{a}}& 0\\ {\displaystyle \frac{-c}{a}}& 1\end{array}\right)A \end{gathered}$$
$$\begin{gathered} \text{We have to make }{\displaystyle \frac{1}{a}}=1 \\ {\text{R}}_2\to a{\text{R}}_2 \\ \left(\begin{array}{cc}1& {\displaystyle \frac{b}{a}}\\ 0& 1\end{array}\right)\text{ = }\left(\begin{array}{cc}{\displaystyle \frac{1}{a}}& 0\\ -c& a\end{array}\right)A \\ \text{We have to make }{\displaystyle \frac{b}{a}}=0 \\ {\text{R}}_1\to {\text{R}}_1-{\displaystyle \frac{b}{a}}{\text{R}}_2 \\ \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\text{ = }\left(\begin{array}{cc}{\displaystyle \frac{1}{a}}-{\displaystyle \frac{b}{a}}(-c)& {\displaystyle \frac{0-b}{a}}\ast a\\ -c& a\end{array}\right)A \end{gathered}$$
$$\begin{gathered} \text{As we know the identity,} \\ I=A{A}^{-1} \\ {A}^{-1}=\left(\begin{array}{cc}{\displaystyle \frac{1+bc}{a}}& -b\\ -c& a\end{array}\right) \\ \text{So,this is the required answer}\text{.} \end{gathered}$$

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