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September 19th, 2016, 09:26 AM  #1 
Newbie Joined: Sep 2016 From: Lisboa Posts: 2 Thanks: 0  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.
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.

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 8a10b7c=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(2y4z) + 3y  2z = 0, rearrange it we get z = 0.7y. substitute it back into x=2y4z, 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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computation, nonzero, solutions 
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