If $A=\left[\begin{array}{cc}1 & 2 \ -5 & 1\end{array} ight]$ and $A^{-1}=x A+y I$, where $I$ is unit matrix of order 2 , then the values of $x$ and $y$ are respectively

Question

If $A=\left[\begin{array}{cc}1 & 2 \ -5 & 1\end{array}ight]$ and $A^{-1}=x A+y I$, where $I$ is unit matrix of order 2 , then the values of $x$ and $y$ are respectively

Aicharya · Accepted Answer

$|\mathrm{A}|=\left|\begin{array}{cc}1 & 2 \ -5 & 1\end{array}ight|=1+10=11$ and adj $\mathrm{A}=\left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight]$ $	herefore \mathrm{A}^{-1}=\frac{1}{11}\left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight]$ Given $\mathrm{A}^{-1}=\mathrm{x} \mathrm{A}+\mathrm{yI}$ $\frac{1}{11}\left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight]=\left[\begin{array}{cc}\mathrm{x} & 2 \mathrm{x} \ -5 \mathrm{x} & \mathrm{x}\end{array}ight]+\left[\begin{array}{ll}\mathrm{y} & 0 \ 0 & \mathrm{y}\end{array}ight]$ $	herefore \frac{1}{11}\left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight] \quad=\left[\begin{array}{cc}\mathrm{x}+\mathrm{y} & 2 \mathrm{x} \ -5 \mathrm{x} & \mathrm{x}+\mathrm{y}\end{array}ight]$ $	herefore 2 \mathrm{x}=\frac{-2}{11} \Rightarrow \mathrm{x}=\frac{-1}{11}$ and $\mathrm{x}+\mathrm{y}=\frac{1}{11} \Rightarrow \mathrm{y}=\frac{2}{11}$

If $A=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $A^{-1}=x A+y I$, where $I$ is unit…

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