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

    Q6.    \begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}

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

                A=IA

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

Use the elementary transformation

             R_1\rightarrow R_1-R_2

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

\Rightarrow          R_2\rightarrow R_2-R_1

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

                R_1\rightarrow R_1-2R_2

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

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

Posted by

seema garhwal

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