TITLE: Higher powers of matrix
QUESTION [5 upvotes]: $$
    A = \begin{bmatrix}
    1 & 1 & 13 \\
    5 & 2 & 6 \\
    -2 & -1 & -3 \\
    \end{bmatrix}
$$
Find $A^{14}+3A-2I$.
One way is to find $A^2$, then $A^4$, then $A^8$, then $A^{14}$.
Another way is using eigen values and diagonal matrix concept.
In our university exam, this question is given only for 5 marks. So I am wondering if there is any simple way to do it.
Appreciate any hint.

REPLY [3 votes]: The characteristic polynomial of $A$ is $\lambda^3+20\lambda+10$, which doesn’t seem to have easily-computable (at least not by hand) roots. Diagonalization or using a decomposition of $A$ into a linear combination of eigenspace projections doesn’t seem like a fruitful way to go.  
An alternative to consider is to use the Cayley-Hamilton theorem to reduce the degree of the polynomial that is to be computed: since $A^3+20A+10I=0$, then $A^{14}+3A-2I$ is equal to its remainder when divided by the characteristic polynomial. This will reduce the degree to at most two and seems like it’ll be less work overall than successively squaring $A$ and then performing a few more matrix multiplications to get to $A^{14}$.  
The long division is tedious, but not difficult, and results in $$A^{14}+3A-2I = 56\,010\,000\,A^2+188\,000\,003\,A+79\,199\,998\,I.$$