I have these three matrices.
$a=\left(\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix}\right)$, $b=\left(\begin{matrix} 0 & 1 \\ -1 & 1 \end{matrix}\right)$, $c=\left(\begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix}\right)$
They belong to $GL_2(\Bbb Q)$. I know the meaning of the period of an element but I don't understand how to verify if these matrices are periodical and eventually how to calculate their period.
A nice approach here is to compute the minimal polynomial (which in this case coincides with the characteristic polynomial) of each matrix. Note that a polynomial $p$ will satisfy $p(M) = 0$ if and only if the minimal polynomial $m(x)$ of $M$ divides $p$.
In this case, we calculate minimal polynomials $$ m_a(x) = x^2 + 1\\ m_b(x) = x^2 - x + 1\\ m_c(x) = x^2 - 2x + 1 $$ $m_a$ and $m_b$ each divide a polynomial of the form $p(x) = x^n - 1$ for some $n$. $m_c$ divides no such polynomial since such a $p(x)$ has no repeated roots over $\Bbb C$, but $m_c$ does.
Alternative approach: trial and error (calculating successive powers) reveals that $a^4 = I$ and $a^6 = I$ (we might expect that this holds for $a$ if we recognize that it is a rotation matrix). However, we find that $$ c^n = \pmatrix{1&-1\\0&1}^n = \left(\pmatrix{1&0\\0&1} + \pmatrix{0&-1\\0&0}\right)^n = \pmatrix{1&0\\0&1} + n \pmatrix{0&-1\\0&0}\\ = \pmatrix{1&-n\\0&1} $$ which makes it clear that if $n > 0$, $c^n$ cannot be the identity matrix.