# Non-trivial solutions

• Sep 19th 2007, 03:10 PM
Thomas
Non-trivial solutions
Find b such that the homogeneous system with augmented matrix below has non-trivial solutions:

$A =
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;-2 & 4 & -18 & 0\;\\\;-4 & 0 & b & 0\;\\\;2 & 1 & 3 & 0\end{vmatrix}
$

So I know a system of equations is homogeneous if all the constant terms are zero, and I know to make a nontrivial solution I need at least one variable to have a nonzero value.

My question is, what is b supposed to be? Is there a single number that it should end up being? Or can it be 1 or 2 or 3, etc.?
• Sep 19th 2007, 03:35 PM
Plato
This is going to be a different approach.
Please investigate what happens to your system if $b = -12$.
After doing so, try to convince yourself that is the answer.
• Sep 19th 2007, 03:47 PM
Thomas
That gives me:

$A =
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & 3 & 0\;\\\;0 & 1 & -3 & 0\;\\\;0 & 0 & 0 & 0\end{vmatrix}
$

So:

$x + 3z = 0$
$y - 3z = 0$

Is that the answer? I'm not getting this.
• Sep 19th 2007, 03:55 PM
Plato
That makes me question if you understand what is going on here?
Is the triple $\left( {3t, - 3t, - t} \right)$ a solution to your system if $t = - 1,1,2, - 2$ or t is any number?
• Sep 19th 2007, 04:00 PM
Thomas
Quote:

Originally Posted by Plato
Is the triple $\left( {3t, - 3t, - t} \right)$ a solution to your system if $t = - 1,1,2, - 2$ or t is any number?

Yes.

I'm wondering where you got $b = -12$ though, that confuses me.
• Sep 19th 2007, 05:24 PM
Plato
Quote:

Originally Posted by Thomas
I'm wondering where you got $b = -12$ though, that confuses me.

EXACTLY THAT! That is what I asked you to find out in the begining.
If you can understand where that comes from then you will understand this problem.
• Sep 19th 2007, 05:49 PM
Thomas
I tried solving the matrix while keeping $b$ in the matrix.

I got:

$A =
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{-b}{4} & 0\;\\\;0 & 1 & \frac{-b - 36}{8} & 0\;\\\;0 & 0 & \frac{5b + 60}{8} & 0\end{vmatrix}
$

I noticed $b = -12$ would make the $\frac{5b + 60}{8} = 0$

Am I getting any closer?
• Sep 20th 2007, 02:22 PM
Thomas
Bump.
• Sep 20th 2007, 02:43 PM
Plato
Quote:

Originally Posted by Thomas
Bump.

Bump, what is this bump?
$b=-12$ makes the rows dependent. What does that imply?
• Sep 20th 2007, 03:27 PM
Thomas
Quote:

Originally Posted by Plato
Bump, what is this bump?
$b=-12$ makes the rows dependent. What does that imply?

I haven't heard the term dependent when doing matrices before, so I don't know.

Could you possibly tell me how you came up with -12 so I understand?
• Sep 21st 2007, 12:38 AM
Opalg
Quote:

Originally Posted by Thomas
I got:

$A =
\begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{-b}{4} & 0\;\\\;0 & 1 & \frac{-b - 36}{8} & 0\;\\\;0 & 0 & \frac{5b + 60}{8} & 0\end{vmatrix}
$

I noticed $b = -12$ would make the $\frac{5b + 60}{8} = 0$

Am I getting any closer?

Yes, you're getting very close. If b = -12 then the bottom row of that matrix becomes all zeros. The corresponding equation is 0 = 0 (which is automatically satisfied, of course), and we're left with only two genuine equations for three unknowns. So you should be able to find a nontrivial solution for them.
• Sep 21st 2007, 04:17 AM
Thomas
Okay, awesome.

When they refer to "trivial" does this just mean the only answer to the equations is 0, and "nontrivial" means the answer can involve numbers other than zero?
• Sep 21st 2007, 04:54 AM
Opalg
Quote:

Originally Posted by Thomas
When they refer to "trivial" does this just mean the only answer to the equations is 0, and "nontrivial" means the answer can involve numbers other than zero?

Yes. A homogeneous system of linear equations always has at least one solution, namely when all the variables are zero. That is called the "trivial" solution. Any other solution is "nontrivial" (even if some of the variables are zero, just so long as they are not all zero).
• Sep 21st 2007, 12:14 PM
Thomas
Okay, thank you for clearing that up. :)