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

    Q5.    \begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}

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A =\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

             R_2\rightarrow R_2-3R_1

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

                  

\Rightarrow          R_1\rightarrow R_1-R_2

\Rightarrow          \begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix} = \begin{bmatrix}4&-1\\-3&1 \end{bmatrix}A

                   R_2\rightarrow R_2-R_1

\Rightarrow          \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix}4&-1\\-7&2 \end{bmatrix}A

\therefore A^{-1}=\begin{bmatrix}4&-1\\-7&2 \end{bmatrix}.

Thus the inverse of matrix A is obtained.

Posted by

seema garhwal

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