Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right],$ show that $A^2-5A+7I=0.$ Use the result to find $A^4$.

Answer 1

$A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right],5A=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]\therefore A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0A^2-5A+7I=0\Rightarrow A^2=5A-7I=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]-\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]\therefore A^4=A^2\cdot A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]=\left[\begin{array}{cc}39& 55\\ -55& -16\end{array}\right]$