Question

The order of the nilpotent matrix $A=\left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]$ is
[a] 2
[b] 3
[c] all multiples of 3
[d] Given matrix is not nilpotent

Answer 1

- Hint: Check if the matrix is not nilpotent by finding the determinant of the given matrix. If the determinant is non-zero, then the matrix is not nilpotent. If the determinant is zero, find ${{A}^{2}},{{A}^{3}},\cdots $ successively and hence find the smallest value of k such that ${{A}^{k}}=O$ and hence find the order of the nilpotent matrix.

Complete step-by-step solution -

Nilpotent matrix: A square matrix A is said to be a nilpotent matrix if there exists $k\in \mathbb{N}$, such that ${{A}^{k}}=O$ where O is a null matrix of the same dimensions as of A. The smallest value of k such that ${{A}^{k}}=O$ is called the order of the nilpotent matrix. Hence if k is the order of the nilpotent matrix, then ${{A}^{n}}=O\forall n\ge k$ and ${{A}^{n}}\ne O\forall n$\det \left( {{A}^{k}} \right)=0$
Since $\det \left( {{A}^{k}} \right)={{\left( \det A \right)}^{k}}$, we get
$\det \left( A \right)=0$
Hence a matrix can be nilpotent if $\det \left( A \right)=0$ and if $\det \left( A \right)\ne 0$ then the matrix cannot be nilpotent.
In the given matrix, we have
$A=\left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]$
Hence $\det \left( A \right)=1\left( -6+6 \right)-1\left( -15+12 \right)+3\left( -5+4 \right)=0+3-3=0$
Since det(A) = 0, the matrix can be a nilpotent matrix.
Now, we have
$A=\left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]$
Hence ${{A}^{1}}\ne O$.
${{A}^{2}}=\left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 0 & 0 \\
   3 & 3 & 9 \\
   -1 & -1 & -3 \\
\end{matrix} \right]$
Since ${{A}^{2}}\ne O$ we have $k\ne 0$
Now, we have ${{A}^{3}}={{A}^{2}}A$
Hence, we have
${{A}^{3}}=\left[ \begin{matrix}
   0 & 0 & 0 \\
   3 & 3 & 9 \\
   -1 & -1 & -3 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 0 & 0 \\
   0 & 0 & 0 \\
   0 & 0 & 0 \\
\end{matrix} \right]=O$
Since ${{A}^{3}}=O$, we have the order of nilpotency = 3.
Hence the order of the given nilpotent matrix is 3, and hence option [b] is correct.

Note: [1] A nilpotent matrix has only 0 as its eigenvalue, and hence characteristic polynomial equation is ${{x}^{n}}=0$. The two statements are equivalent, i.e. A matrix is nilpotent if ${{x}^{n}}=0$ is the characteristic polynomial equation of the matrix and if the matrix is nilpotent, then ${{x}^{n}}=0$ is the characteristic polynomial equation.
The characteristic polynomial equation of a matrix A is the polynomial equation $\det \left( A-xI \right)=0$.
Hence we can prove that $A=\left[ \begin{matrix}
   1 & 1 & 3 \\
   5 & 2 & 6 \\
   -2 & -1 & -3 \\
\end{matrix} \right]$ is a nilpotent matrix by verifying $\det \left( A-xI \right)=0$ is the polynomial equation ${{x}^{3}}=0$