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If $A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $A^{-1}=x A+y I$, where $I$ is unit matrix of order 2 , then the values of $x$ and $y$ are respectively
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$\frac{-1}{11}, \frac{2}{11}$
$|\mathrm{A}|=\left|\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right|=1+10=11$ and adj $\mathrm{A}=\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$
$\therefore \mathrm{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$
Given $\mathrm{A}^{-1}=\mathrm{x} \mathrm{A}+\mathrm{yI}$
$$
\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]=\left[\begin{array}{cc}\mathrm{x} & 2 \mathrm{x} \\ -5 \mathrm{x} & \mathrm{x}\end{array}\right]+\left[\begin{array}{ll}\mathrm{y} & 0 \\ 0 & \mathrm{y}\end{array}\right]
$$
$\therefore \frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right] \quad=\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & 2 \mathrm{x} \\ -5 \mathrm{x} & \mathrm{x}+\mathrm{y}\end{array}\right]$
$\therefore 2 \mathrm{x}=\frac{-2}{11} \Rightarrow \mathrm{x}=\frac{-1}{11}$ and $\mathrm{x}+\mathrm{y}=\frac{1}{11} \Rightarrow \mathrm{y}=\frac{2}{11}$
$\therefore \mathrm{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]$
Given $\mathrm{A}^{-1}=\mathrm{x} \mathrm{A}+\mathrm{yI}$
$$
\frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right]=\left[\begin{array}{cc}\mathrm{x} & 2 \mathrm{x} \\ -5 \mathrm{x} & \mathrm{x}\end{array}\right]+\left[\begin{array}{ll}\mathrm{y} & 0 \\ 0 & \mathrm{y}\end{array}\right]
$$
$\therefore \frac{1}{11}\left[\begin{array}{cc}1 & -2 \\ 5 & 1\end{array}\right] \quad=\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & 2 \mathrm{x} \\ -5 \mathrm{x} & \mathrm{x}+\mathrm{y}\end{array}\right]$
$\therefore 2 \mathrm{x}=\frac{-2}{11} \Rightarrow \mathrm{x}=\frac{-1}{11}$ and $\mathrm{x}+\mathrm{y}=\frac{1}{11} \Rightarrow \mathrm{y}=\frac{2}{11}$
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