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Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

    Q4.    \begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}

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A=\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

             R_2\rightarrow R_2-2R_1

\Rightarrow          \begin{bmatrix} 2 &3 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A

                  

\Rightarrow          R_1\rightarrow R_1-3R_2

\Rightarrow          \begin{bmatrix} -1 &0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}A

               R_2\rightarrow R_2+R_1

\Rightarrow          \begin{bmatrix} -1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}7&-3\\5&-2\end{bmatrix}A

                     R_1\rightarrow -R_1

\Rightarrow          \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}-7&3\\5&-2\end{bmatrix}A

                     

\therefore A^{-1}=\begin{bmatrix}-7&3\\5&-2\end{bmatrix}

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Posted by

seema garhwal

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