Cramer rule

By using Cramer rule solve the following systems of equations:

Variant 1
\(\mathbf A\mathbf x=\mathbf b\) over \(\mathbb R\) for

\( \mathbf A= \begin{pmatrix} 1 & 2 & -2 & 2 \\ 2 & 3 & 1 & 1 \\ -1 & 2 & 3 & -1 \\ 1 & -1 & 1 & -1 \\ \end{pmatrix} \) and \( \mathbf b = \begin{pmatrix} -2 \\ 2 \\ -1 \\ 4 \end{pmatrix} \).
Hint
If \(\mathbf A\) is regular, then by Cramer rule the system \(\mathbf A\mathbf x=\mathbf b\) has solution \(x_i=\frac{|\mathbf A_{i\to \mathbf b}|}{|\mathbf A|}\) where the matrix \(\mathbf A_{i\to \mathbf b}\) becomes from \(\mathbf A\) by replacement of the \(i\)-th column by the vector \(\mathbf b\).
Answer
\(|\mathbf A|=6, \mathbf x=(12/6,-6/6{,}6/6{,}0/6)^T=(2,-1{,}1,0)^T\).
Variant 2
over \(\mathbb Z_5\):

\[\begin{eqnarray*} 2x + y + 4z &=& 1 \\ 3x + y + 4z &=& 2 \\ 2x + 4y + 2z &=& 3 \end{eqnarray*}\]
Answer
\(|A|=4=4^{-1}, (x,y,z)^T=(4/4{,}1/4{,}0/4)^T=(1{,}4,0)^T\).