Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$ Show that $A^2-5A+71=0$, Use this to find $A^4$.

Answer 1

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

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

$=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$

$5A=5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right];7I=7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A^2-5A+7I$

$=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

$=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$=0$ - proved

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

or, $A^2=5A-7I$

$\therefore A^2.A^2=5A^2-7IA.A$

or, $A^4=5A^2-7A^2=-2A^2$

or, $A^4=-2\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]=\left[\begin{array}{cc}-16& -10\\ 10& -6\end{array}\right]$