Question : Verify Cayley Hamilton theorem for the following matrix
\[A=\left(\begin{matrix}1&-1&1\\2&-1&0\\1&0&0\end{matrix}\right) \]
Solution :
Let
\[A=\left(\begin{matrix}1&-1&1\\2&-1&0\\1&0&0\end{matrix}\right)\]
Then Characteristics equation of A is
\[\left|A-xI\right|=0\]
\[\left|\begin{matrix}1-x&-1&1\\2&-1-x&0\\1&0&-x\end{matrix}\right|=0\]
\[\left(1-x\right)\left(-x\left(-1-x\right)-0\right)-\left(-1\right)\left(-2x-0\right)+1\left(-\left(-1-x\right)\right)=0\]
\[x-x^3-2x+1+x=0\]
\[x^3-1=0 \]
In order to verify Caley Hamilton theorem for the matrix A we shall prove that
\[A^3-I=0 \\\ (1)\]
Where I is a 3 by 3 identity matrix. for the given matrix
\[ A^2=\left(\begin{matrix}1&-1&1\\2&-1&0\\1&0&0\end{matrix}\right)\left(\begin{matrix}1&-1&1\\2&-1&0\\1&0&0\end{matrix}\right)=\left(\begin{matrix}0&0&1\\0&-1&2\\1&-1&1\end{matrix}\right)\]
\[A^3=A^2A=\left(\begin{matrix}0&0&1\\0&-1&2\\1&-1&1\end{matrix}\right)\left(\begin{matrix}1&-1&1\\2&-1&0\\1&0&0\end{matrix}\right)\]
\[=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right)=I\]
\[A^3-I=0 \\\\ (2)\]
Since eqns (1) and (2) are same.
This verifies Caley Hamilton theorem for the given matrix.