Question

Show that $\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$ is nilpotent matrix of order $3$

Answer 1

A matrix $A$ is said to be nilpotent matrix of order $m$ provided it satisfies the relation $A^k=0$ and $A^{k-1}\ne 0$ where $k$ is a positive integer

Let $A=\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$

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

$=\left[\begin{array}{ccc}1+5-6& 1+2-3& 3+6-9\\ 5+10-12& 5+4-6& 15+12-18\\ -2-5+6& -2-2+3& -6-6+9\end{array}\right]$

$=\left[\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right]$

$A^3=A^2.A=\left[\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right]\times \left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$

$=\left[\begin{array}{ccc}0+0+0& 0+0+0& 0+0+0\\ 3+15-18& 3+6-9& 9+18-27\\ -1-5+6& -1-2+3& -3-6+9\end{array}\right]=0$

$\therefore A^3=0$ where $k=3$

Hence, the matrix $A$ is nilpotent of order $3$