Find $A^5-4A^4+7A^3+11A^2-A-10I$

Question

If$$A= \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix},$$then find $A^5-4A^4+7A^3+11A^2-A-10I$ where $I$ is Identity matrix of $2^{nd}$ order. Answer should come in terms of $A$ and $I$.

My approach: I thought it would might end up in a pattern so I found

$$A^2=\begin{bmatrix} 9 & 16 \\ 8 & 17 \end{bmatrix}$$ and similarly $A^3$ but got no such pattern. I don't want to evaluate it to up to $A^5$. Any better approach or solution is much appreciated.

Answer 1

As the eigenvalues of $A$ are $5$ and $-1$ you have $A^2 -4A - 5I = 0_{2\times2}$

Polynomial division gives:

• $A^5-4A^4+7A^3+11A^2-A-10I = (A^3+12A+59I)(A^2-4A-5I)+295A + 285I$

So, the result is $$\boxed{295A + 285I}$$

Answer 2

The characteristic polynomial of that matrix is $\lambda^2-4\lambda-5$. So, by the Hamilton-Cayley theorem, $A^2-4A-5\operatorname{Id}=0$. In other words, $A^2=4A+5\operatorname{Id}$. Can you take it from here?