# Solve the following non-homogeneous system of linear equation by determinant method : $2x-y+z=2\;,6x-3y+3z=6\;,4x-2y+2z=4$

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
$\Delta = \begin{vmatrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{vmatrix} = -\begin{vmatrix} 2 & 1 & 1 \\ 6 & 3 & 3 \\ 4 & 2 & 2 \end{vmatrix} = 0 \: ( \because C_2 \equiv C_3)$
The equations may or may not be consistent.
Step 2
$\Delta_x = \begin{vmatrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{vmatrix} = 0 ( = \Delta)$
$\Delta_y = \begin{vmatrix} 2 & 2 & 1 \\ 6 & 6 & 3 \\ 4 & 4 & 2 \end{vmatrix} = 0 ( \because C_1 \equiv C_2)$
$\Delta_z = \begin{vmatrix} 2 & -1 & 2 \\ 6 & -3 & 6 \\ 4 & -2 & 4 \end{vmatrix} = 0 ( \because C_1 \equiv C_3)$
$\Delta = \Delta_x =\Delta_y=\Delta_z=0$
All 2nd order minors of $\Delta, \Delta_x, \Delta_y, \Delta_z$ vanish. $\Delta$ has atleast one nonzero element. The system reduces to one equation. It is consistent with infinitely many solutions. Let $y = s, \: z = t, \: and\: t \in R$.
Then $x = \large\frac{2+s-t}{2}$ from the first equation.
$(x,y,z) =\bigg( \large\frac{2+s-t}{2}, s, t \bigg) \: \: s,t \in R$

edited May 30, 2013