<s>[INST] Solving a system of equation modul0 5 Consider the system of linear equations $$\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3\\ 1 \end{pmatrix}  $$
a) Solve the system in $\mathbb{F}_5$
I just want to make sure my solution is correct:
We have: $$A=\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\\ \Rightarrow A^{-1}= \frac{1}{42}\begin{pmatrix} 6 & 3\\ -2 & 6 \end{pmatrix}\\ \equiv \begin{pmatrix} 3 & \frac{3}{2}\\ -1 & 3 \end{pmatrix} \mbox{ } [5]$$
Therefore: $\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3 & \frac{3}{2}\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 3\\ 1 \end{pmatrix}= \begin{pmatrix} \frac{21}{2}\\ 0 \end{pmatrix} [5]   $
 [/INST] Note that modulo $5$ one has
$$
A =
\begin{bmatrix} 6 & -3\\ 2 & 6 \end{bmatrix}
=
\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}.
$$
The latter matrix has determinant $-3 \equiv 2$, whose inverse is $3$, so 
$$
A^{-1} =
\begin{bmatrix} 3 & -1\\ -1 & 3 \end{bmatrix}.
$$
Now
$$
\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}
=
A^{-1}
\begin{bmatrix} 3\\ 1 \end{bmatrix}
=
\begin{bmatrix} 3 & -1\\ -1 & 3 \end{bmatrix}.
\begin{bmatrix} 3\\ 1 \end{bmatrix}
=
\begin{bmatrix} 3\\ 0 \end{bmatrix},
$$
which is your result, as
$$
\frac{21}{2} \equiv 21 \cdot 3 \equiv 3 \pmod{5}.
$$
</s>