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

    Q9.    \begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}

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A=\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

             R_1\rightarrow R_1-R_2

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

\Rightarrow          R_2\rightarrow R_2-2R_1

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

                  R_1\rightarrow R_1-3R_2

\Rightarrow          \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}A

Thus using elementary transformation the inverse of A is obtained as

\therefore A^{-1}=\begin{bmatrix}7&-10\\-2&3 \end{bmatrix}.

Posted by

seema garhwal

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