Question

Find the order of the matrix $\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$
$\therefore A^2=A.A=\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]\times \left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$
$=\left[\begin{array}{ccc}1+5-6& 1+2-3& 3+6-9\\ 5+10-12& 5+4-6& 15+12-18\\ -2-5+6& -2-2+3& -6-6+9\end{array}\right]$
$=\left[\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right]$
$\therefore A^3=A^2.A=\left[\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right]\times \left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$
$=\left[\begin{array}{ccc}0+0+0& 0+0+0& 0+0+0\\ 3+15-18& 3+6-9& 9+18-37\\ -1-5+6& -1-2+3& -3-6+9\end{array}\right]$
$=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=0$
$\therefore A^3=0,i.e.a^k=0$

Here $k=3$. Hence, $A$ is nilpotent matrix of order $3$.