Question

Find $A^2-5A+6IIfA=\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]$

Answer 1

Here, $A^2=A\times A=\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]=\left[\begin{array}{ccc}4+0+1& 0+0-1& 2+0+0\\ 4+2+3& 0+1-3& 2+3+0\\ 2-2+0& 0-1+0& 1-3+0\end{array}\right]=\left[\begin{array}{ccc}5& -1& 2\\ 9& -2& 5\\ 0& -1& -2\end{array}\right]$
$\therefore A^2-5A+6I=\left[\begin{array}{ccc}5& -1& 2\\ 9& -2& 5\\ 0& -1& -1\end{array}\right]-5\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]+6\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}5& -1& 2\\ 9& -2& 5\\ 0& -1& -1\end{array}\right]-\left[\begin{array}{ccc}10& 0& 5\\ 10& 5& 15\\ 5& -5& 0\end{array}\right]+\left[\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right]=\left[\begin{array}{ccc}5-10+6& -1+0+0& 2-5+0\\ 9-10+0& -2-5+6& 5-15+0\\ 0-5+0& -1+5+0& -2+0+6\end{array}\right]=\left[\begin{array}{ccc}1& -1& -3\\ -1& -1& -10\\ -5& 4& 4\end{array}\right]$