Question

Find the characteristic equation of the matrix

A= [ 1 2 3
4 1 -1
-3 2 5 ]
and verify Cayley-Hamilton theorem for it.

Answer 1

The characteristic equation of the given matrix $A$ is $det(A-\lambda I)=0$, where $\lambda$ is any scalar.

$\Rightarrow \left | \begin{matrix} 1-\lambda & 2 & 3 \\ 4 & 1-\lambda & -1 \\ -3 & 2 & 5-\lambda \end{matrix} \ \right |=0$

$\Rightarrow (1-\lambda )\left [ (1-\lambda )(5-\lambda )+2 \right ]-2\left [ 4(5-\lambda )-3 \right ]+3\left [ 8+3(1-\lambda ) \right ]=0$

$\Rightarrow (1-\lambda )(\lambda ^{2}-6\lambda +7)-2(17-4\lambda )+3(11-3\lambda )=0$

$\Rightarrow -\lambda ^{3}+7\lambda ^{2}-14\lambda +6=0$

Therefore, the characteristic of the given matrix is given by

$\lambda ^{3}-7\lambda ^{2}+14\lambda -6=0$

Now let us verify the Cayley- Hamilton theorem.

The matrix equation is given by

$A ^{3}-7A ^{2}+14A -6I=O$

$A^2=\left [ \begin{matrix} 1 & 2 & 3 \\ 4 & 1 & -1 \\ -3 & 2 & 5 \end{matrix} \ \right ]\left [ \begin{matrix} 1 & 2 & 3 \\ 4 & 1 & -1 \\ -3 & 2 & 5 \end{matrix} \ \right ]=\left [ \begin{matrix} 0 & 10 & 16 \\ 11 & 7 & 6 \\ -10 & 6 & 14 \end{matrix} \ \right ]$

And

$A^3=\left [ \begin{matrix} 0 & 10 & 16 \\ 11 & 7 & 6 \\ -10 & 6 & 14 \end{matrix} \ \right ]\left [ \begin{matrix} 1 & 2 & 3 \\ 4 & 1 & -1 \\ -3 & 2 & 5 \end{matrix} \ \right ]=\left [ \begin{matrix} -8 & 42 & 70 \\ 21 & 41 & 56 \\ -28 & 14 & 34 \end{matrix} \ \right ]$

Now

$A ^{3}-7A ^{2}+14A -6I$

$\left [ \begin{matrix} -8 & 42 & 70 \\ 21 & 41 & 56 \\ -28 & 14 & 34 \end{matrix} \ \right ]-\left [ \begin{matrix} 0 & 70 & 112 \\ 77 & 49 & 42 \\ -70 & 42 & 98 \end{matrix} \ \right ]$

$+\left [ \begin{matrix} 14 & 28 & 42 \\ 56 & 14 & -14 \\ -42 & 28 & 70 \end{matrix} \ \right ]-\left [ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix} \ \right ]$

$=\left [ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \ \right ]$

$=O$

Therefore, $A ^{3}-7A ^{2}+14A -6I=O$

Hence, Cayley-Hamilton Theorem is verified.