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Solve the following non-homogeneous system of linear equation by determinant method : $ 2x+y-z=4\;,x+y-2z=0\;,3x+2y-3z=4 $

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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} 2 & 1 & -1 \\ 1 & 1 & -2 \\ 3 & 2 & -3 \end{vmatrix} =2(-3+4)-1(-3+6)-1(2-3)$
$ = 2-3+1=0$
$ \Delta = 0. \: \: \: \therefore$ the system may or maynot be consistent.
Step 2
$ \Delta_x = \begin{vmatrix} 4 & 1 & -1 \\ 0 & 1 & -2 \\ 4 & 2 & -3 \end{vmatrix} = 4(-3+4)+4(-2+1)$
$ = 4-4=0$
$ \Delta_y = \begin{vmatrix} 2 & 4 & -1 \\ 1 & 0 & -2 \\ 3 & 4 & -3 \end{vmatrix} = 2(0+8)-4(-3+6)-1(4-0)$
$ = 16-12-4=0$
$ \Delta_z = \begin{vmatrix} 2 & 1 & 4 \\ 1 & 1 & 0 \\ 3 & 2 & 4 \end{vmatrix} = 2(4-0)-1(4-0)+4(2-3)$
$ = 8-4-4=0$
$ \Delta = \Delta_x = \Delta_y = \Delta_z = 0$
Consider the 2 x 2 minor $ \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} $ of $ \Delta \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = 2-1=1 \neq 0.\: \: \: \therefore$ there is atleast one nonzero minor.
$ \therefore $ the system is consistent with infinitely many solutions.
Step 3
It reduces to two equations.
$ 2x+y-z=4$
$\: \: x+y-2z=0$
Let $ z = k$ where $ k \in R$
Then
$ 2x+y = 4+k$
$ \: \: x+y = +2k$
$ \Delta = \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix}=2-1 = 1 \neq 0, \: \Delta_x = \begin{vmatrix} 4+k & 1 \\ +2k & 1 \end{vmatrix} = 4+k-2k = 4-k$
$ \Delta_y = \begin{vmatrix} 2 & 4+k \\ 1 & +2k \end{vmatrix} = +4k-4-k = -4+3k$
By cramers rule
$ x = \large\frac{\Delta x}{\Delta} = 4-k, \: y = \large\frac{\Delta y}{\Delta} = -4+3k$
$ \therefore (x,y,z) = (4-k,\: -4+3k, \: k) k \in R$
answered May 28, 2013 by thanvigandhi_1
 

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