Given are $3$ vectors in $\mathbb{Z}_{3}^{3}$. Decide if these are linearly independent by using rule of Sarrus.
$$v_{1}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, v_{2}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, v_{3}=\begin{pmatrix} 0\\ 1\\ -1 \end{pmatrix}$$
I need to change in $v_{3}$ the $-1$ because $-1 \text{ mod }3 = 2$. So here is the first thing I'm not sure about. Do I have to use Sarrus first and then change numbers (in case they don't equal 0,1 or 2) or rather change the number first and continue with Sarrus?
Here I first change number and I'm not sure if that's ok.
$$v_{1}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, v_{2}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, v_{3}=\begin{pmatrix} 0\\ 1\\ 2 \end{pmatrix}$$
Write this as $3 \times 3$ matrix and write it like Sarrus (write additionally $v_{1}$ and $v_{2}$):
Now we need to calculate the determinant $D$ and if it doesn't equal zero then the vectors are linearly independent.
$$D=1 \cdot 0 \cdot 2+1 \cdot 1 \cdot 1+ 0 \cdot 1 \cdot 0-1 \cdot 0 \cdot 0- 0 \cdot 1 \cdot 1 - 2 \cdot 1 \cdot 1$$
$$D=0+1+0-0-0-2=-1$$
And thus the vectors are linearly independent.
Is everything done correctly here? Because we write an exam soon and I need to know if I can do it like this please.