Question

Let $A=\left[\begin{array}{cc}2& 3\\ -1& 5\end{array}\right]$. If $A^{-1}=xA+yI$, find $x+2y$.

Answer 1

$A=\left[\begin{array}{cc}2& 3\\ -1& 5\end{array}\right]$

$A^{-1}=xA+yI$

$\Rightarrow A^{-1}A=xAA+yIA$

$\Rightarrow I=x{A}^2+yA$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=x\left[\begin{array}{cc}2& 3\\ -1& 5\end{array}\right]\left[\begin{array}{cc}2& 3\\ -1& 5\end{array}\right]+y\left[\begin{array}{cc}2& 3\\ -1& 5\end{array}\right]$

$=x\left[\begin{array}{cc}1& 21\\ -7& 22\end{array}\right]+y\left[\begin{array}{cc}2& 3\\ -1& 5\end{array}\right]$

$comparingelementsonbothsides$

$x+2y=1and-7x-y=0$

$solvingaboveequationsgives$

$x=\frac{-1}{13}andy=\frac{7}{13}$

$\Rightarrow x+2y=1$