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November 6th, 2016, 12:31 PM  #1 
Newbie Joined: Nov 2016 From: Croatia Posts: 4 Thanks: 0  [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) 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 
[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], $X_1 = a + 3(1/10  7b) + 4(5b) = 3/10  a  b$. 
November 6th, 2016, 02:29 PM  #3 
Newbie Joined: Nov 2016 From: Croatia Posts: 4 Thanks: 0  
November 6th, 2016, 05:09 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
Yes.


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