<s>[INST] Compute $A^5-4A^4-7A^3+11A^2-A-10I$ for $A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$ by Cayley-Hamilton Question: Use Cayley-Hamilton theorem to express $A^5-4A^4-7A^3+11A^2-A-10I$, where $$A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$$
Attempt: I only know how to get the characteristic polynomial equation which is $\lambda^2-4\lambda -5$ for that matrix. By definition $A^2-4A-5=0$. But where does that  5 degree polynomial come from?
 [/INST] Continuing your analysis,
$$A^2 - 4A -5I=0\implies A^2 = 4A+5I$$
Thus,
$$\begin{array}{cc}

A = A\\
A^2 = 4A + 5I\\
A^3 = 4A^2 + 5A &= 4(4A+5I) + 5A = 21A +20I\\
\vdots
\end{array}$$
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