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September 19th, 2016, 09:26 AM   #1
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Question 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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September 19th, 2016, 01:14 PM   #2
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It is confusing. x=y=z=0 is a solution no matter what a,b,c are.
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September 20th, 2016, 08:59 AM   #3
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But that is not the answer, the answer is

8a-10b-7c=0

Do you know how to get there?
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September 21st, 2016, 05:34 PM   #4
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With the system in matrix form, note that the determinant is 8a - 10b - 7c.
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October 1st, 2016, 02:34 PM   #5
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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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