Matrix and eigenvectors 
$\quad$The matrix $\mathbf A=\frac19\begin{bmatrix}
7   & -2  &  0 \\
-2  & 6   &  3  \\
0   & 2   &  5   \\
\end{bmatrix}$ has eigenvalues $1$, $\frac23$ and $\frac13$n with the corresponding eigenvectors $ \mathbf v_1=\begin{bmatrix}
-2                  \\
2                    \\
1                     \\
\end{bmatrix}$, $\mathbf{v}_2=\begin{bmatrix}
2                       \\
1                        \\
2                         \\
\end{bmatrix}$, $\mathbf v_3=\begin{bmatrix}
1                           \\
2                            \\
-2                            \\
\end{bmatrix}$.                
$\text{(i)}$ Find the general solution of the equation $\mathbf x_{k+1}=\mathbf{Ax}_k$ if $\mathbf x_0=\begin{bmatrix}
1                                  \\
11                                  \\
-2                                   \\
\end{bmatrix}$.

For this question, does it mean I should find out what $x_k$ is in terms of $A$? 
Can anyone help me please?

You have
$$A=P^{-1}DP$$
With
$$P=\begin{bmatrix}
-2&2&1\\
2&1&2\\
1&2&-2\\
\end{bmatrix}$$
and
$$D=\begin{bmatrix}
1&0&0\\
0&2/3&0\\
0&0&1/3\\
\end{bmatrix}$$
Then
$$x_n=A^nx_0= P^{-1}DP \; P^{-1}DP \dots P^{-1}DP =P^{-1}D^nP \,x_0$$
With
$$D^n=\begin{bmatrix}
1&0&0\\
0&(2/3)^n&0\\
0&0&(1/3)^n\\
\end{bmatrix}$$
After simplification of the matrix product, and noticing $P^2=9I_3$, thus $P^{-1}=\frac19P$,
$$x_n=\frac1{3^n}\begin{bmatrix}
2^{n+1}-4\cdot3^{n}+3\\
4\cdot3^{n}+2^{n}+6\\
2\cdot\left(3^{n}+2^{n}-3\right)\\
\end{bmatrix}$$
