A matrix raised to a high power ($87$) So, I have this matrix: $$\pmatrix {0&0&0&-1\\1&0&0&0\\0&1&0&0\\0&0&1&0}^{87}$$
My teacher never discussed eigenvalues. So, I do not know what they are and there must be another way to do this (without multiplying the matrix $87$ times). 
Thanks for your help.

Note that your matrix acts like the permutation $\sigma=(2341)$, i.e. $1 \mapsto 2$, $2 \mapsto 3$, $3 \mapsto 4$, $4 \mapsto 1$ such that each time it spits a minus sign to the fourth, third, second and first column respectively.
The reason for this is that you can think of a matrix in the following way:


*

*The first column of a matrix is where $e_1$ goes.

*The second column of a matrix is where $e_2$ goes.


And so on, so forth. 
Looking at the problem in this way helps us directly compute any power of $A$ easily. In fact, we have the following closed form for $A^n$:
$$A^n=\pmatrix{(-1)^{\lfloor \frac{n+8}{4} \rfloor}
e_{\sigma^n(1)} && (-1)^{\lfloor \frac{n+9}{4} \rfloor}e_{\sigma^n(2)} && (-1)^{\lfloor \frac{n+10}{4} \rfloor}e_{\sigma^n(3)} && (-1)^{\lfloor \frac{n+3}{4} \rfloor}e_{\sigma^n(4)}}
$$
Where $e_n$ is the $n$-th standard basis vector, $\sigma=(2341)$ and $\lfloor \cdot \rfloor$ is the floor function.
In particular, for $n=87$, we get:
$$A^{87}=\pmatrix{(-1)^{\lfloor \frac{95}{4} \rfloor}
e_{\sigma^{87}(1)} && (-1)^{\lfloor \frac{96}{4} \rfloor}e_{\sigma^{87}(2)} && (-1)^{\lfloor \frac{97}{4} \rfloor}e_{\sigma^{87}(3)} && (-1)^{\lfloor \frac{90}{4} \rfloor}e_{\sigma^{87}(4)}}
$$
Now, since $\sigma^4=e$, we have $\sigma^{87}=\sigma^{84}\sigma^{3}=(\sigma^{4})^{21}\sigma^{3}=\sigma^3$
We only need to calculate: $\sigma^3(1)=4, \sigma^3(2)=1, \sigma^3(3)=2, \sigma^3(4)=3$
Therefore, our answer in the closed form is:
$$A^{87}=\pmatrix{-
e_{4} && +e_{1} && +e_{2} && +e_{3}}
$$
And we can expand it to see that our final answer is:
$$A^{87}=\pmatrix{
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
-1&0&0&0}
$$
