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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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