Solve the following non-homogeneous system of linear equation by determinant method : $3x+y-z=2\;,2x-y+2z=6\;,2x+y-2z=-2$

Toolbox:
• Consistency of a system of 2 ( or 3 ) linear equations in 2 ( or 3 ) unknowns. Case 1 : If $\Delta \neq 0$ thesystem is consistent and has a unique solution. Cramer's rule can be used. Case 2 : If $\Delta = 0$, there are two / ( four) possibilities:
• 2a: If $\Delta = 0$, and atleast one of the values $\Delta_ x, \Delta_ y ( or \: \Delta_ z )$ is nonzero, the system is inconsistent and has no solution.
• 2b: If $\Delta = 0\: and \: \Delta_x=\Delta_y ( = \Delta_z)=0$ and at least one element ( second order minor ) of $\Delta$ is nonzero, the system is consistent with infinitely many solution. The system reduces to one equation ( two equations )
• 2c: If $\Delta = 0$ and $\Delta_x=\Delta_y = \Delta_z=0$ and all their 2 x 2 minors are 0, and $\Delta$ has atleast one nonzero element, then the system is consistent with infinitely many solutions. The system reduces to one equation.
• 2d: If $\Delta = 0, \Delta_x=\Delta_y = \Delta_z=0$, all 2 x 2 minors of $\Delta$ are zero but there is at least one 2 x 2 minor of $\Delta_x, \Delta_y \: or \: \Delta_z$ that is nonzero, then the system is inconsistent and has no solution.
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
$x=\large\frac{\Delta x}{\Delta} = \large\frac{4}{4}=1, \: y = \large\frac{\Delta y}{\Delta} = \large\frac{8}{4}=2, \: \: z = \large\frac{\Delta z}{\Delta} = \large\frac{12}{4}=3$
$(x,y,z) = (1,2,3)$