Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$ ,show that $A^2-5A+7$ $I=0$ and hence find $A^{-1}$.

Answer 1

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

$A^2-5A+7I=0$

$A^{-1}{A}^2-5A^{-1}A+7A^{-1}I=0$ by multiplying by $A^{-1}$

$\Rightarrow A-5I+7A^{-1}=0$ since $A{A}^{-1}=I$

$\Rightarrow 7A^{-1}=-A+5I$

$\Rightarrow A^{-1}=\frac{1}{7}[5I-A]$

$=\frac{1}{7}\left[\begin{array}{cc}5(\frac{10}{01})& -(\frac{31}{-12})\end{array}\right]$

$=\frac{1}{7}\left[\begin{array}{cc}(\frac{50}{05})& -(\frac{31}{-12})\end{array}\right]$

$\frac{1}{7}=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{cc}2/7& -1/7\\ 1/7& 3/7\end{array}\right]$