Question

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

Answer 1

Let $A=\left[\begin{array}{cc}8& 1\\ -1& 2\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
Then $A^2=\left[\begin{array}{cc}3& 1\\ 3& 1\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$
$=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$
$A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-5\left[\begin{array}{cc}-3& 1\\ -1& 2\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$
$=\left[\begin{array}{cc}8-15+7& 5-5+10\\ -5+5+0& 8-10+7\end{array}\right]$
$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$