Linearizing a Matrix Power

A shorter method would be to write $A^2=m\,A+n\,I$ .

Then find $A^2$ by simple multiplication and equate it to $m\,A+n\,I$ to find $m=3$ $n=10$ .

Now write $A^3=A\left(A^2\right)=A\left(3A+10I\right)=19A+30I$ .

Similarly find remaining powers of $A$ as above while substituting $A^2$ every time with $3A+10I$ to finally arrive at $A^6=2223A+4510I$ . So $\alpha=2223$ $\beta=4510$ and $\alpha+\beta=\boxed{6733}$ .

NOTE: This solution still may not provide the best process yet

A more general approach to this problem is using the Cayley-Hamilton Theorem .

The characteristic polynomial for $A$ is $p(\lambda)=det(A-\lambda I)=(-1-\lambda)(4-\lambda)-6=\lambda^{2}-3\lambda-10$

Applying Cayley-Hamilton, we have $p(A)=A^{2}-3A-10I=0$ , which can be rearranged to $A^2=3A+10I$

By the repeated process of multiplying both sides by $A$ and substituting $A^2=3+10I$ , we eventually find $\alpha$ and $\beta$ :

$A^{3}=(3A+10I)A$

$=3A^{2}+10A$

$=3(3A+10I)+10A$

$=19A+30I$

Similarly, $A^{4}=19(3A+10I)+30A=87A+190I$ , $A^{5}=87(3A+10I)+190A=451A+870I$ , and $A^{6}=451(3A+10I)+870A=2223A+4510I$

Therefore, $\alpha=2223$ and $\beta=4510$ , so $\alpha+\beta=\boxed{6733}$

A very simple way to do this is to consider an eigenvector of A call it V, with eigenvalue x. The characteristic poly of A is x^2-3x-10. So the eigenvalues are 5 and -2. Now we know A^6 V= aAV+bIV. Now since V is an eigenvector with eigenvalue x we have A^6V=x^6V and AV=xV. So we have an equation x^6V=axV+bV. When we factor out V we get (x^6)V=(ax+b)V. So we have x^6=ax+b we sub in 5 and -2 and get two linear equations which we solve to get a+b=6733.

Rewrite the equation as: ${{A}^{6}}-\alpha A=\beta I$ And calculate the values: $\left[ \begin{matrix} 2287+\alpha & 4446-2\alpha \\ 6669-3\alpha & 13402-4\alpha \\ \end{matrix} \right]=\left[ \begin{matrix} \beta & 0 \\ 0 & \beta \\ \end{matrix} \right]$

That leads to: $\alpha =2223;\,\beta =4510$