<s>[INST] Matrix to power 30 using eigen values issue 
$$ \text{If matrix } A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \text{ then find } A^{30}.$$

I tried to approach through diagonalization using eigen values method.
I got eigen values as $-1, 1, 1$
As per diagonalization $ A = P*D*P^{-1}.$ So $ A^{30} = P*D^{30}*P^{-1}. $
But $ D^30 = I.$
So, $ A^{30} = P*I*P^{-1} = I $
But $ A^{30} $ is not equal to I. If we do general multiplication without all these. 
Where is the mistake in my approach?
 [/INST] You have computed the eigenvalues of $A$ to be $\{-1, 1, 1\}$.  The repeated eigenvalue may be an obstacle to diagonalization.  In this case, the geometric and algebraic multiplicities of the eigenvalue $1$ are different, so A is not diagonalizable.  You continued as if $A$ were diagonalizable, so this is the mistake in your approach.
You have added a follow-on question in comments (instead of to your question, so you should not be surprised if other answers do not address it).  I see that others have demonstrated Jordan normal form and induction.  Another method is binary decomposition of the exponent and repeated squaring to get power-of-$2$ powers of $A$: \begin{align*}
A^1 &= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}  \text{,} \\
A^2 &= A^1 \cdot A^1 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}  \text{,} \\
A^4 &= A^2 \cdot A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}  \text{,} \\
A^8 &= A^4 \cdot A^4 = \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 4 & 0 & 1 \end{pmatrix}  \text{,} \\
A^{16} &= A^8 \cdot A^8 = \begin{pmatrix} 1 & 0 & 0 \\ 8 & 1 & 0 \\ 8 & 0 & 1 \end{pmatrix}  \text{,} \\
A^{30} = A^{11110_{\,2}} &= A^{16}\cdot A^8 \cdot A^4 \cdot A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 15 & 1 & 0 \\ 15 & 0 & 1 \end{pmatrix}  \text{.}
\end{align*}
Of course, after computing $A^4$ or $A^8$, perhaps one would notice the pattern...
There is a slightly faster way to get there using the above method.  $A^{-1}$ is easy enough to compute (using the minors method, for example).  $$
A^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}
$$  Then, $A^{30} = (A^{15})^2 = (A^{16} \cdot A^{-1})^2$.  This replaces three multiplies with a multiply and an inverse.  Continuing to think about this leads to the computationally intractable problem of optimal addition chains.
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