Determine the $n$-th power of matrix Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1  \\ }$$ May you help me with the answer since the book states :  $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n &  3\cdot4^n & 2+4^n  \\ }$$ while mine misses the ${1\over 6}$. So who is right?
$$$$Thanks in advance for any help you are able to provide.

I get for the symbolic form depending on n and the symbolic eigenvalues $\alpha,\beta,\gamma$ (where we insert later the known values $\small \alpha=0,\beta=1,\gamma=4$)
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 \cdot \beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & -4\cdot\beta^n+3\cdot\alpha^n+\gamma^n & \\
 -2\cdot\beta^n+2\cdot\gamma^n & 3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n-3\cdot\alpha^n+\gamma^n \\
 -2\cdot\beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & 2\cdot\beta^n+3\cdot\alpha^n+\gamma^n
 \end{array}
$$
and replacing with the actual values for $\small \alpha,\beta,\gamma$ we get
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & -4 +3\cdot 0^n+4^n & \\
 -2+2\cdot 4^n & 3\cdot 0^n+3\cdot 4^n & 2-3\cdot 0^n+4^n \\
 -2+2\cdot 4^n & -3\cdot 0^n+3\cdot 4^n & 2+3\cdot 0^n+4^n
 \end{array}
$$
For $n \in \mathbb N \text{ and } n \gt 0$ this reduces to
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2\cdot 4^n & 3\cdot 4^n & -4+4^n & \\
 -2+2\cdot 4^n & 3\cdot 4^n & 2+4^n \\
 -2+2\cdot 4^n & 3\cdot 4^n & 2+4^n
 \end{array}
$$
$ \qquad \qquad $ which matches the solution of the book.      
For $n=0$ this gives with $0^0=1$ 
$$ A^0 = \frac16 \small \begin{array} {|r|r|r|r}
 4 +2 & -3+3 & -4 +3+1 & \\
 -2+2 & 3+3 & 2-3+1 \\
 -2+2 & -3+3 & 2+3+1
 \end{array} = \frac16 \small \begin{array} {|r|r|r|r}
 6 & 0 & 0 & \\
 0 & 6 & 0 \\
 0 & 0 & 6
 \end{array} = I
$$
(So the textbook-solution should be commented with the required $0^0$-definition)

I used the diagonalization-procedure of Pari/GP for this and to get a textform for this answer:
M = [2,2,0;1,2,1;1,2,1]
tmpM=mateigen(M)
tmpW=tmpM^-1
tmpD=tmpW*M*tmpM
MPow = tmpM*matdiagonal(['a,'b,'c])*tmpW 
\\ on text-level we have later to replace a by \alpha^n and so on

