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Applications of Matrices and Determinants
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Solve the following non-homogeneous system of linear equation by determinant method: $2x+3y=5\;,4x+6y=12 $
tnstate
class12
bookproblem
ch1
sec-1
exercise1-4
p35
q2
sec-a
easy
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asked
Mar 29, 2013
by
poojasapani_1
retagged
Sep 19, 2013
by
vijayalakshmi_ramakrishnans
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1 Answer
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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 & 3 \\ 4 & 6 \end{vmatrix} = 12-12=0$
The system may or may not be consistent.
Step 2
$ \Delta x = \begin{vmatrix}5 & 3 \\ 12 & 6 \end{vmatrix} = 30-36 = -6 \neq 0$
$ \Delta = 0$ but $ \Delta x \neq 0.\: \: \therefore$ the system is inconsistent and has no solution.
answered
May 27, 2013
by
thanvigandhi_1
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