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November 6th, 2016, 12:31 PM   #1
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[College Matrices] Can't figure it out at 5th try.

I did everything, but I always get the wrong answer, I mean, maybe the solutions that's been given to me is wrong, but that's highly unlikely. Can someone review it for me?

EDIT: The correct answer should be: (x1, x2, x3, x4, x5) = (3/10 - a - b, a, -1/2 - 20b, 1/10 - 7b, 5b)
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Last edited by stipanrelix; November 6th, 2016 at 12:34 PM.
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November 6th, 2016, 02:10 PM   #2
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[1] $\ \ \ X_1 + X_2 + 0X_3 - 3X_4 - 4X_5 = 0$
[2] $\ \ \ X_1 + X_2 -\ \ X_3 + 2X_4 -\ \, X_5 = 1$
[3] $\ 2X_1 + 2X_2 + X_3 - \,\,X_4 + 3X_5 = 0$

[1] + [2] - [3]: $-2X_3 - 8X_5 = 1$, so $X_3 = -1/2 - 4X_5$
3[1] - [2] - [3]: $-10X_4 - 14X_5 = -1$, so $X_4 = 1/10 - (7/5)X_5$

Now let $X_5 = 5b$ (so that $X_3 = -1/2 - 20b$ and $X_4 = 1/10 - 7b$).
Also, let $X_2 = a$.
From [1], $X_1 = -a + 3(1/10 - 7b) + 4(5b) = 3/10 - a - b$.
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November 6th, 2016, 02:29 PM   #3
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Quote:
Originally Posted by skipjack View Post
Now let $X_5 = 5b$ (so that $X_3 = -1/2 - 20b$ and $X_4 = 1/10 - 7b$).
Also, let $X_2 = a$.
From [1], $X_1 = -a + 3(1/10 - 7b) + 4(5b) = 3/10 - a - b$.
So I actually did it correctly, I just assigned a different parameter?
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November 6th, 2016, 05:09 PM   #4
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Yes.
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