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State Cayley-Hamilton theorem & verify the same for $A=\begin{bmatrix}1&3\\2&2\end{bmatrix}$
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Cayley-Hamilton theorem states that every square matrix satisfies its characteristics equation.

The characteristic equation is

$\begin{bmatrix}1-\lambda&3\\2&2-\lambda\end{bmatrix}=0$

After simplification

$(1-\lambda)(2-\lambda)-6=0$

$2-\lambda-2\lambda+\lambda^2-6=0\\\lambda^2-3\lambda-4=0$

Cayley-Hamilton Theorem states that this equation satisfied by the matrix A i.e

$A^2-3A-4I=0$

$A^2= \begin{bmatrix}1&3\\2&2\end{bmatrix}\begin{bmatrix}1&3\\2&2\end{bmatrix}$

$=\begin{bmatrix}7&9\\6&10\end{bmatrix}$

$A^2-3A-4I= \begin{bmatrix}7&9\\6&10\end{bmatrix}-\begin{bmatrix}3&9\\6&6\end{bmatrix}-\begin{bmatrix}4&0\\0&4\end{bmatrix}$

$=\begin{bmatrix}0&0\\0&0\end{bmatrix} \text { Hence verified }$