How to solve a system of equations over a finite field? I need to solve a system of equations over $\mathbb{Z}_{11}$.
My system is:
$$ \left\{ 
\begin{array}{l}
2x + 5y + z = 8 \\
7x + 6y + 8z = 10 \\
10x + 3y + 4z = 6
\end{array}
\right.
$$
In matrix form:
$$
\begin{bmatrix}
2 & 5 & 1\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix} = 
\begin{bmatrix}
8\\
10\\
6\\
\end{bmatrix}
$$
after $R_1*6$ :
$$
\begin{bmatrix}
1 & 8 & 6\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{bmatrix}= 
\begin{bmatrix}
4\\
10\\
6\\
\end{bmatrix}
$$
$R_2-7*R_1$ and $R_3-10*R_1$:
$$
\begin{bmatrix}
1 & 8 & 6\\
0 & 5 & 10\\
0 & 0 & 1\\
\end{bmatrix}= 
\begin{bmatrix}
4\\
4\\
10\\
\end{bmatrix}
$$
$R_2*9$ :
$$
\begin{bmatrix}
1 & 8 & 6\\
0 & 1 & 2\\
0 & 0 & 1\\
\end{bmatrix}= 
\begin{bmatrix}
4\\
3\\
10\\
\end{bmatrix}
$$
$$\left\{ 
\begin{array}{l}
 z = 10 \\
y = 3 - 2z = 3 - 20 = -17=5 \\
x = 4 - 8y - 6z = 4 - 8*5 - 6*10 = -96 = 3
\end{array}
\right.$$
But after substitution found values in the initial system I have an error:
$$ \left\{ 
\begin{array}{l}
2x + 5y + z = 2*3 + 5*5+10= 41 = 8 \\
7x + 6y + 8z = 7*3 + 6*5 + 8*10 = 131 = 10 \\
10x + 3y + 4z = 10*3 + 3*5 + 4*10= 85 = 8 \neq 6
\end{array}
\right.
$$
Do I have an arithmetic error or the whole way is wrong? Is there an easier and faster way to solve it?

You made a calculation error when calculating $R_3-10* R_1$. The last row in the matrix should be $[0,0,10|10]$ rather than $[0,0,1|10]$. With this fix, you get that the solution to the system is $x=y=z=1$, which you can easily check actually solves the system.
