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

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  • 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} 1 & 2 & 1 \\ 3 & 3 & -1 \\ 2 & 1 & -2 \end{vmatrix}$
$ = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 2 & 1 & -2 \end{vmatrix} R_2 \rightarrow R_2-R_3$
$= 0\: (\therefore R_1 \equiv R_3)$
$ \Delta = 0.$ The system may or may not be consistent.
Step 2
$ \Delta_x = \begin{vmatrix} 6 & 2 & 1 \\ 3 & 3 & -1 \\ -3 & 1 & -2 \end{vmatrix}$
$= \begin{vmatrix}  6 & 2 & 1 \\ -3 & 1 & -2 \\ -3 & 1 & -2 \end{vmatrix} R_2 \rightarrow R_2-R_1$
$=0$
$ \Delta_y = \begin{vmatrix}  1 & 6 & 1 \\ 3 & 3 & -1 \\ 2 & -3 & -2 \end{vmatrix}$
$= \begin{vmatrix}  1 & 6 & 1 \\ 2 & -3 & -2 \\ 2 & -3 & -2 \end{vmatrix} R_2 \rightarrow R_2-R_1$
$=0$
$ \Delta_z = \begin{vmatrix}  1 & 2 & 6 \\ 3 & 3 & 3 \\ 2 & 1 & -3 \end{vmatrix}$
$ = \begin{vmatrix}  1 & 2 & 6 \\ 2 & 1 & -3 \\ 2 & 1 & -3 \end{vmatrix} R_2 \rightarrow R_2-R_1$
$ =0$
$ \Delta = \Delta_x = \Delta_y = \Delta_z = 0$
Consider the 2nd order minor of $ \Delta \begin{vmatrix} 1 & 2 \\ 3 & 3 \end{vmatrix} = 3-6=-3 \neq 0$
$ \therefore$ the system is consistent with infinitely many solutions. It reduces to two equations.
$ x+2y+z=6\: \: \: \: \: \: \: 3x+3y-z=3$
Let $z = k \in R$ Then
$ x+2y=6-k$
$3x+3y=3+k$
$ \Delta = \begin{vmatrix}1 & 2 \\ 3 & 3 \end{vmatrix} = 3-6=-3. $ By cramer's rule
$ \Delta_x= \begin{vmatrix} 6-k & 2 \\ 3+k & 3 \end{vmatrix} = 18-3k-6-2k = 12-5k$
$ \Delta_y = \begin{vmatrix}1 & 6-k \\ 3 & 3+k \end{vmatrix} = 3+k-18+3k = -15+4k$
$ x = \large\frac{\Delta x}{\Delta} = \large\frac{12-5k}{-3}= \large\frac{5k-12}{3}$
$ y = \large\frac{\Delta y}{\Delta} = \large\frac{-15+4k}{-3}= \large\frac{15-4k}{3}$
$ (x,y,z) = \bigg( \large\frac{5k-12}{3}, \large\frac{15-4k}{3}, k \bigg) \: k \in R$

 

answered May 28, 2013 by thanvigandhi_1
edited May 28, 2013 by thanvigandhi_1
 

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