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]   $

This method is legitimate since, if a solution exists, the determinant is non-zero mod $5$, and thus is invertible mod $5$.
To continue, $21 \equiv 1 \pmod 5$ and $\frac{1}{2}=2^{-1} \equiv 3 \pmod 5$.
So the solution becomes $$\begin{bmatrix} 3 \\ 0 \\ \end{bmatrix}.$$
