I have trouble solving the second question that is to compute for $A^4$ without just brute forcing. Is there another way aside from that? I found that I could use a Jordan Decomposition for large powers of a matrix but seeing that this is a 3×3 matrix I don't know if I am just overcomplicating things. My prof. was said to give questions that have hidden hints or shortcuts and I am just not so confident that what I am doing is right.
$$\text{Question:}$$ Let $A= \begin{bmatrix} 6 & -6 & 6 \\ -2 & 5 & -3 \\-4 & 6 & -4\end{bmatrix}$. Verify that $A^2-5A+6I=0$, where I is the $3 \times 3$ identity matrix, and compute $A^4$.
You'd use Cayley-Hamilton theorem. It's something along the lines of 'given the characteristic polynomial P($\lambda$) = det(A - $\lambda$I) = 0, then the matrix A satisfies the matrix equation P(A) = 0, where you just put the matrix in the space of $\lambda$ and insert the identity matrix where there's no $\lambda$. The way you'd solve your question is along the lines of, do the above. You can factorise the matrix equation (which will be at most at 3rd order in A) as due to the rules of matrix multiplication (i.e. A, B, C are nxn matrices and b,c constants, then A(bB + cC) = bAB + cAC ecc.) this will be allowed. Then show that you can 'divide' by the element you've factored. This should yield your equation for $aA^{2} + bA + cI = 0$, and should agree with what you want. Then just take $A^{4} = A^{2}A^{2}$ and then expand the equation, appropriately substituting $A^{2} = \frac{-b}{a}A + \frac{-c}{a}I$. Then it will just be a case of addition and multiplying by constants to get $A^{4}$.
EDIT
To find the n-th power, use the ideas I stated to construct equations for the n-th power of A.