Cayley Hamilton theorem : If A be any square matrix over a field of real or complex numbers, then it satisfies it’s own characteristics equation.
Example : 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 Cayley 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 Cayley Hamilton Theorem for the given matrix A