Question

If $A=\left[\begin{array}{cc}3& 5\\ -2& 3\end{array}\right]$
Find $A^{-1}$ if this matrix satisfies the equation
$A^2-6A+19I=0$.

Answer 1

Given, $A^2-6A+19I=0$

Multiply with $A^{-1}$ on both sides,

$A^{-1}{A}^2-6A^{-1}A+19A^{-1}I=0$

$A-6I+19A^{-1}=0$

$A^{-1}=\frac{1}{19}(-A+6I)$

$A^{-1}=\frac{1}{19}(-\left[\begin{array}{cc}3& 5\\ -2& 3\end{array}\right]+6\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])$

$A^{-1}=\frac{1}{19}\left[\begin{array}{cc}3& -5\\ 2& 3\end{array}\right]$