Say we want to solve the system \begin{eqnarray*} 3x + 2y + z -2w&=&4\\ 2x - y + 2z - 5w &=& 15\\ 4x + 2y \phantom{+ 2x} - 5w&=&1\\ 3x \phantom{+ 2y} - 2z -4w &=& 1. \end{eqnarray*} The associated matrix is $$\begin{pmatrix} 3 & 2 & 1 & -2\\ 2 & -1 & 2 & -5\\ 4 & 2 & 0 & -1 \\ 3 & 0 & -2 & -4 \end{pmatrix}$$

whose determinant is $\Delta=-65$ . Since the determinant is non-zero, we can use Cramer's rule. To obtain the value of the $k$ -th variable, we replace the $k$ -th column of the matrix above by the column vector $$\begin{pmatrix} 4\\ 15\\ 1\\ 1\end{pmatrix},$$ the determinant of the obtained matrix is divided by $\Delta$ and the resulting value is the wanted solution.

So $$ x=\frac{\Delta_1}{\Delta}= \frac{ \begin{vmatrix} 4 & 2 & 1 & -2\\ 15 & -1 & 2 & -5\\ 1 & 2 & 0 & -1\\ 1 & 0 & -2 & -4 \end{vmatrix} }{-65}=\frac{-65}{-65}=1 $$

$$ y=\frac{\Delta_2}{\Delta}= \frac{ \begin{vmatrix} 3 & 4 & 1 & -2\\ 2 & 15 & 2 & -5\\ 4 & 1 & 0 & -1\\ 3 & 1 & -2 & -4 \end{vmatrix} }{-65}=\frac{130}{-65}=2 $$

$$ z=\frac{\Delta_3}{\Delta}= \frac{ \begin{vmatrix} 3 & 2 & 4 & -2\\ 2 & -1 & 15 & -5\\ 4 & 2 & 1 & 1\\ 3 & 0 & 1 & -4 \end{vmatrix} }{-65}=\frac{-195}{-65}=3 $$

$$ w=\frac{\Delta_4}{\Delta}= \frac{ \begin{vmatrix} 3 & 2 & 1& 4 \\ 2 & -1 & 2 & 15\\ 4 & 2 & 0 & 1 \\ 3 & 0 & -2 & 1 \end{vmatrix} }{-65}=\frac{65}{-65}=-1 $$