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Matrix Questions- not related to hsc (1 Viewer)

mathsfreak1010

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Find all values of m for which
x+2y+mz=-1
2x+y-z=3
mx-2y+z=1
has a unique solution.
In the cases where there is no unique solutions, solve the system.

The answer is m can equal all values apart from -1 and -5
i get why is cant be -1 but i dont get how to m cant equal -5???


THANKS IN ADVANCE
 

HeroicPandas

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I made a mistake while reducing the matrix down into row-echelon form..

I'll assume that you did it, could you please tell me the last row the matrix? Or entries 3,3 and 3,4?
 

HeroicPandas

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Matrix after Gaussian Elimination: http://snag.gy/jVmfx.jpg (scroll all the way down, line (5) and pretend there is a line separating columns 2 and 3)

For this matrix to have a unique solution (Because if m^2 /3 + 2m +5/3 = 0, you won't be able to solve for z and so no unique sol'n for z)

m^2 /3 + 2m +5/3 =/= 0
m^2 +6m + 5 =/= 0
(m+5)(m+1) =/= 0
m =/= -5 or m =/= -1 (WHICH ONE?)?

Look at the entry in the last and last column, -7/3 -7m/3

Suppose that m = -5, then m^2 /3 + 2m +5/3 = 0 and -7/3-7m/3 =/=0
Our last will be 0 0 0 | constant........is this possible? No! It means no solutions. So we conclude that m=/=-5

Suppose that m = -1, then m^2 /3 + 2m +5/3 = 0 and -7/3-7m/3 = 0
Our last column will be 0 0 0 | 0, and so we disregard it and inspect row 2...you will notice that you can't solve for z and you must let it be a parameter which leads to infinitely many solutions which means NO UNIQUE and so m=/=-1
 
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Drongoski

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For unique solution, the associated matrix should be non-singular

i.e. its determinant = -(m^2 +6m+5) = -(m+1)(m+5) is nonzero

Since the quadratic = 0 for m = -5, -1 you have the required answer (viz m =/= -1, -5).
 
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HeroicPandas

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For unique solution, the associated matrix should be non-singular

i.e. its determinant = -(m^2 +6m+5) = -(m+1)(m+5) is nonzero

Since the quadratic = 0 for m = -5, -1 you have the required answer (viz m =/= -1, -5).
Beautiful
 

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