Step 1

$ \Delta = \begin{vmatrix} 3 & 1 & -1 \\ 2 & -1 & 2 \\ 2 & 1 & -2 \end{vmatrix} =3(2-2)-1(-4-4)-1(2+2)$

$ = 0+8-4=4 \neq 0$

Cramer's rule can be used to obtain the unique solution.

Step 2

$ \Delta_x = \begin{vmatrix} 2 & 1 & -1 \\ 6 & -1 & 2 \\ -2 & 1 & -2 \end{vmatrix} = 2(2-2)-1(-12+4)-1(6-2)$

$ = 0+8-4=4$

$ \Delta_y =\begin{vmatrix} 3 & 2 & -1 \\ 2 & 6 & 2 \\ 2 & -2 & -2 \end{vmatrix} = 3(-12+4)-2(-4-4)-1(-4-12)$

$ -24+16+16=8$

$ \Delta_z = \begin{vmatrix} 3 & 1 & 2 \\ 2 & -1 & 6 \\ 2 & 1 & -2 \end{vmatrix} = 3(2-6)-1(-4-12)+2(2+2)$

$ = -12+16+8=12$

Step 3

$ x=\large\frac{\Delta x}{\Delta} = \large\frac{4}{4}=1, \: y = \large\frac{\Delta y}{\Delta} = \large\frac{8}{4}=2, \: \: z = \large\frac{\Delta z}{\Delta} = \large\frac{12}{4}=3$

$ (x,y,z) = (1,2,3)$