How to prove a matrix property? Let $$A=\begin {bmatrix}
                    1 &amp; 3 &amp;  1\\
                    1 &amp; 1 &amp; -2\\
                    1 &amp; 2 &amp; -1\\
                   \end{bmatrix}
                  $$

I  need to demonstrate this relation:
$$\forall n \in \mathbb{N}, A^n \neq I_3$$
where $I_3$=identity matrix.
If you have seen eigenvalues, you know that if $\lambda$ is an eigenvalue of $A$, 
$\lambda^n$ is an eigenvalue of $A^n$. But all eigenvalues of $A$ are such that $|\lambda| \neq 1$ ($\lambda \approx 1.839$ and $\lambda \approx -0.420 \pm 0.606 i$), thus it is impossible, whatever $n$, to have $|\lambda|^n=1$ (the eigenvalue of $I_3$).
Proof using mathematical induction:

The base case: for $n=1$ we have $$A^1 = \begin {bmatrix}
                    1 &amp; 3 &amp;  1\\
                    1 &amp; 1 &amp; -2\\
                    1 &amp; 2 &amp; -1\\
                   \end{bmatrix}$$
So $A^1 \ne I$ is true for the base case.

Induction hypothesis: Assume that $A^n \ne I$ is a correct mathematical proposition and let proof that $A^{n+1} \ne I$ is also correct.
$$A^{n+1}=A^{n}\cdot A$$
According to the base case we have $A \ne I$ and according to the induction hypothesis we have $A^n \ne I$ , so $$A^n\cdot A \ne I \cdot I$$
$$\Leftrightarrow A^{n+1} \ne  I$$

Hence $A^{n+1} \ne  I$ is true. So $$\forall n \in \mathbb{N}: A^n\ne I$$