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

    Q14.    \begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}

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

                A=IA

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

             R_2\rightarrow \frac{R_2}{2}

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

\Rightarrow          R_1\rightarrow R_1-R_2

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

Hence, we can see all upper values of matrix are zeros in L.H.S  so A^{-1} does not exist. 

Posted by

seema garhwal

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