Find large power of a non-diagonalisable matrix 
If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$.

The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresponding to eigenvalue $-1$. So, it is not diagonalizable.
Is there any other way to compute the power?

Since the given matrix is rather simple, you could also compute a few powers:
\begin{align*}
A^1&=
\begin{pmatrix}
  1 & 0 & 0 \\
  1 & 0 & 1 \\
  0 & 1 & 0
\end{pmatrix}\\
A^2&=
\begin{pmatrix}
  1 & 0 & 0 \\
  1 & 1 & 0 \\
  1 & 0 & 1
\end{pmatrix}\\
A^3&=
\begin{pmatrix}
  1 & 0 & 0 \\
  2 & 0 & 1 \\
  1 & 1 & 0
\end{pmatrix}\\
A^4&=
\begin{pmatrix}
  1 & 0 & 0 \\
  2 & 1 & 0 \\
  2 & 0 & 1
\end{pmatrix}\\
A^5&=
\begin{pmatrix}
  1 & 0 & 0 \\
  3 & 0 & 1 \\
  2 & 1 & 0
\end{pmatrix}\\
A^6&=
\begin{pmatrix}
  1 & 0 & 0 \\
  3 & 1 & 0 \\
  3 & 0 & 1
\end{pmatrix}\\
\end{align*}
Can you notice some kind of pattern and prove it by induction? (Or, if you prefer, you can do the same thing with $B=A^2$ and try to find general form of $B^n$.)

Of course, this is "naive" approach - without using any theory you have learned about matrices. For more complicated matrices it would be rather difficult to spot some kind of pattern. That's why methods which work for arbitrary matrices are more useful.
