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

Q11. $\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$

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$A=\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-3R_1$

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

$\Rightarrow$ $R_2\rightarrow R_2+R_1$

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

$R_1\rightarrow -R_1$

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

$R_2\rightarrow \frac{R_2}{-2}$

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

thus the inverse of matrix A is

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

Posted by

seema garhwal

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