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Compute the indicated power of the matrix: $A^8$

$ A = \begin{bmatrix}2&1&2\\2&1&2\\2&1&2\end{bmatrix} $

I calculated the eigenvalues:

$ \lambda_1 = \lambda_2 = 0, \lambda_3 = 5 $

and the corresponding eigenvectors/spaces:

$ v_1 = \begin{bmatrix}-1\\2\\0\end{bmatrix}, v_2 = \begin{bmatrix}-1\\0\\1\end{bmatrix}, v_3 = \begin{bmatrix}1\\1\\1\end{bmatrix} $

Using the eigenvectors to construct P and the eigenvalues to construct D:

$ P = \begin{bmatrix}-1&-1&1\\2&0&1\\0&1&1\end{bmatrix}, D = \begin{bmatrix}0&0&0\\0&0&0\\0&0&5\end{bmatrix}$

Now I can calculate $ A^8 = PD^8P^{-1}$

However how do I calculate $P^{-1}$? The textbook does not give an example of how it calculated the inverse and it seems to come out of nowhere. The only way to calculate the inverse that i know of is to use the adjoint and determinant of P:

$P^{-1} = \frac{adj(P)}{det(|P|)} $

Which is rather time consuming so I thought that they might have used another way to do this. Any suggestions?

Edit: Sorry I think my question is not clear. Although I have to compute the power of a matrix, my problem is with the inverse of $P$. I have to use the method i described above. I'm not necessarily interested in the answer of $A^8$ but more on how I can calculate the inverse of P

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You can find $P^{-1}$ by row reducing $[P|I]$ to $[I|P^{-1}]$. – user84413 Mar 16 at 16:51
    
@Macavity, I need the inverse of P, not A... the column space of P is linearly independent – Tadzio Mar 16 at 19:11
1  
Finding the inverse seems overkill here, but you could do Gauss Jordan method (already suggested above). Check en.m.wikipedia.org/wiki/… – Macavity Mar 16 at 20:11
up vote 2 down vote

Hint
Use $A^3=5A^2$ to express $A^8$ in terms of $A, A^2$.

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To understand why $A^3 = 5A^2$ you may wish to look up the Cayley–Hamilton theorem. – jim Mar 16 at 17:16
1  
If you're not familiar with it, check en.m.wikipedia.org/wiki/Cayley–Hamilton_theorem. The matrix satisfies its own characteristic polynomial, ie $\lambda^3-5\lambda^2=0$. Hence $A^8=(A^3A)^2=(5A^2A)^2=5^2(A^3)^2=5^4(A^2)^2=5^4A^4=....=5^7A$. – Macavity Mar 16 at 17:22
up vote 1 down vote

this is a rank one matrix, therefore we can write $A = bc^\top$ where $b = (1,1,1)^\top, c= (2,1,2).$ now, $A^8 = (cb^\top)^7bc^\top = 5^7\pmatrix{2&1&2\\2&1&1\\2&1&2}.$

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