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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}} \\ 

\]


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

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