 My Math Forum [College Matrices] Can't figure it out at 5th try.
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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. November 6th, 2016, 02:10 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216  $\ \ \ X_1 + X_2 + 0X_3 - 3X_4 - 4X_5 = 0$  $\ \ \ X_1 + X_2 -\ \ X_3 + 2X_4 -\ \, X_5 = 1$  $\ 2X_1 + 2X_2 + X_3 - \,\,X_4 + 3X_5 = 0$  +  - : $-2X_3 - 8X_5 = 1$, so $X_3 = -1/2 - 4X_5$ 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 , $X_1 = -a + 3(1/10 - 7b) + 4(5b) = 3/10 - a - b$. November 6th, 2016, 02:29 PM   #3
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Quote:
 Originally Posted by skipjack Now let $X_5 = 5b$ (so that $X_3 = -1/2 - 20b$ and $X_4 = 1/10 - 7b$). Also, let $X_2 = a$. From , $X_1 = -a + 3(1/10 - 7b) + 4(5b) = 3/10 - a - b$.
So I actually did it correctly, I just assigned a different parameter? November 6th, 2016, 05:09 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Yes. Tags 5th, college, figure, matrices Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post stipanrelix Linear Algebra 1 November 6th, 2016 02:32 PM Al_Ch Elementary Math 2 May 21st, 2009 05:53 PM fibonaccilover New Users 2 December 31st, 1969 04:00 PM

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