If the matrix $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \ -5 & 1\end{array} ight]$ and $\mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}$, when $I$ is a unit matrix of order 2 , then the value of $2 x+3 y$ is

Question

If the matrix $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \ -5 & 1\end{array}ight]$ and $\mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}$, when $I$ is a unit matrix of order 2 , then the value of $2 x+3 y$ is

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{A}=\left[\begin{array}{cc}1 & 2 \ -5 & 1\end{array}ight] \ 	herefore \quad & \mathrm{A} \mid=11 \ & \mathrm{~A}^{-1}=\frac{1}{|\mathrm{~A}|}\left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight]=\frac{1}{11}\left[\begin{array}{cc}1 & -2 \ 5 & 1\end{array}ight] \ & \mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}, 	ext { we get } \ & {\left[\begin{array}{cc}\frac{1}{11} & \frac{-2}{11} \ \frac{5}{11} & \frac{1}{11}\end{array}ight]=\left[\begin{array}{cc}x & 2 x \ -5 x & x\end{array}ight]+\left[\begin{array}{ll}y & 0 \ 0 & y\end{array}ight] }\end{aligned}$ $\begin{array}{ll}	herefore & {\left[\begin{array}{cc}\frac{1}{11} & \frac{-2}{11} \ \frac{5}{11} & \frac{1}{11}\end{array}ight]=\left[\begin{array}{cc}x+y & 2 x \ -5 x & x+y\end{array}ight]} \ \Rightarrow & x=\frac{-1}{11} 	ext { and } y=\frac{2}{11} \ 	herefore & 2 x+3 y=2\left(\frac{-1}{11}ight)+3\left(\frac{2}{11}ight)=\frac{4}{11}\end{array}$

If the matrix $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $\mathrm{A}^{-1}=x…

Solution