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September 19th, 2016, 09:26 AM   #1
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Nonzero solutions computation

Hi, I do not know how to start this question and what the question is asking for. Thank you so much! Please help.
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Last edited by skipjack; October 2nd, 2016 at 12:48 AM.

 September 19th, 2016, 01:14 PM #2 Global Moderator   Joined: May 2007 Posts: 6,756 Thanks: 696 It is confusing. x=y=z=0 is a solution no matter what a,b,c are. Thanks from topsquark
 September 20th, 2016, 08:59 AM #3 Newbie   Joined: Sep 2016 From: Lisboa Posts: 2 Thanks: 0 But that is not the answer, the answer is 8a-10b-7c=0 Do you know how to get there?
 September 21st, 2016, 05:34 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond With the system in matrix form, note that the determinant is 8a - 10b - 7c.
 October 1st, 2016, 02:34 PM #5 Newbie   Joined: Oct 2016 From: London Posts: 1 Thanks: 0 from 2nd row, x - 2y + 4z = 0, rearrange it we get x = 2y- 4z. substitute it into the 1st row, 2(2y-4z) + 3y - 2z = 0, rearrange it we get z = 0.7y. substitute it back into x=2y-4z, we get x=-0.8y. ok, now sub both x=-0.8y and z=0.7y into 3rd row, we have -0.8ay + by+ 0.7cy=0 y(-0.8a+b+0.7c)=0 Since y is nonzero as given in the question, (-0.8a+b+0.7c) must be zero. -0.8a+b+0.7c=0 multiply the whole expression by 10, -8a+10b+7c=0

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