Question

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

• A. 2
• B. 3
• C. All the multiples of 3
• D. Given matrix is not a nilpotent matrix

Answer 1

The correct option is B 3

Let $A=\left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$
$A^2=A\times 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]$$=\left[\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right]$
$\therefore A^3=A^2\times A=\left[\begin{array}{ccc}0& 0& 0\\ 3& 3& 9\\ -1& -1& -3\end{array}\right]\times \left[\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ -2& -1& -3\end{array}\right]$$=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=O$
$\therefore A^3=O$
Hence,A is nilpotent of order 3.
Hence option B is correct .