# Trivial/Non Trivial Solutions

• Oct 15th 2008, 03:46 PM
Hellreaver
Trivial/Non Trivial Solutions
Suppose that you are given a matrix:

0 3 1
0 3 1
2 0 0
2 0 -1

How do you determine whether the solution is trivial (and independent) or nontrivial (and dependent)?
• Oct 15th 2008, 04:43 PM
ThePerfectHacker
Quote:

Originally Posted by Hellreaver
Suppose that you are given a matrix:

0 3 1
0 3 1
2 0 0
2 0 -1

How do you determine whether the solution is trivial (and independent) or nontrivial (and dependent)?

What solution?

This is Mine 1:)9:):)th Post!!!
• Oct 15th 2008, 04:48 PM
Hellreaver
Thats one of my problems. I'm told to find out whether the vectors are linearly independent or not, and the textbook tells me that if the linear system has non trivial solutions, it is dependent, and if it has trivial solutions, it will be dependent. Am I supposed to find row echelon form? And if so, what do I do with it?
• Oct 15th 2008, 05:02 PM
ThePerfectHacker
I think the problem is asking to define $A = \begin{bmatrix} 0 & 3 & 1 \\ 0& 3 & 1 \\ 2&0&0 \\ 2&0&-1 \end{bmatrix}$
And then determine if $A\bold{x} = \bold{0}$ has non-trivial solutions.
Where $\bold{x} \in \mathbb{R}^3$ and $\bold{0} = \begin{bmatrix}0&0&0&0\end{bmatrix}^T$.

Since the number of equations exceede the number of variables the homogenous system will have non-trivial solutions.
If you want to find those solutions you need to use Gaussian-Jordan elimination.
Do you know how to do that?
• Oct 15th 2008, 05:02 PM
Scopur
I think they want you to augment it with zeros, and see if it has a trivial/non-trivial solution. And if its for linear independence you can see the first and 2nd equations are redundant and therefore not L.I. So it is dependent.
• Oct 15th 2008, 05:12 PM
Hellreaver
Quote:

Originally Posted by ThePerfectHacker
I think the problem is asking to define $A = \begin{bmatrix} 0 & 3 & 1 \\ 0& 3 & 1 \\ 2&0&0 \\ 2&0&-1 \end{bmatrix}$
And then determine if $A\bold{x} = \bold{0}$ has non-trivial solutions.
Where $\bold{x} \in \mathbb{R}^3$ and $\bold{0} = \begin{bmatrix}0&0&0&0\end{bmatrix}^T$.

Since the number of equations exceede the number of variables the homogenous system will have non-trivial solutions.
If you want to find those solutions you need to use Gaussian-Jordan elimination.
Do you know how to do that?

I have an idea how to do it...
But what about where the number of equations matches the number of variables:
ie
0 3 1 4
0 3 1 4
2 0 0 2
2 0 -1 1

In this case, what do I need to do to determine linear dependence?
• Oct 15th 2008, 05:17 PM
ThePerfectHacker
Quote:

Originally Posted by Hellreaver
In this case, what do I need to do to determine linear dependence?

You set up the system of equations in the form to do Gaussian-Jordan elimination.
Then you find the row reduced echelon form of the matrix.
Once you do that it should be clear if the system has only a trivial solution of not*.

*)To get only trivial solutions the row reduced echelon form must be $\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0& 0&0&1&0\end{bmatrix}$
Because this matrix leads to the following equations:
$1x_1+0x_2+0x_3+0x_4 = 0$
$0x_1+1x_2+0x_3+0x_4 = 0$
$0x_1+0x_2+1x_3+0x_4 = 0$
$0x_1+0x_3+0x_3+1x_3 = 0$
Which immediately means $x_1=x_2=x_3=x_4=0$.
Therefore, it has only the trivial solutions.
• Oct 15th 2008, 05:32 PM
Hellreaver
I didn't get this matrix:
$\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0& 0&0&1&0\end{bmatrix}$

My matrix has a row of zeroes prior to reduced row echelon form...

$\begin{bmatrix}0&3&1&4&0\\0&3&1&4&0\\2&0&0&2&0\\2& 0&-1&1&0\end{bmatrix}$

Which then becomes:
$\begin{bmatrix}2&0&0&2&0\\0&3&1&4&0\\0&3&1&4&0\\2& 0&-1&1&0\end{bmatrix}$

And then:

$\begin{bmatrix}1&0&0&1&0\\0&3&1&4&0\\0&3&1&4&0\\0& 0&-1&-1&0\end{bmatrix}$

~ $\begin{bmatrix}1&0&0&1&0\\0&1&1/3&4/3&0\\0&3&1&4&0\\0&0&-1&-1&0\end{bmatrix}$

~ $\begin{bmatrix}1&0&0&1&0\\0&1&1/3&4/3&0\\0&0&0&0&0\\0&0&-1&-1&0\end{bmatrix}$

What does this mean in relation to its linear independence?
• Oct 15th 2008, 05:43 PM
Scopur
He wasnt saying that $\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0& 0&0&1&0\end{bmatrix}
$
was the answer.. he was saying that if you get this as your answer then your vectors would be linearly independent. The row of zeros indicates that that specific vector was a linear combination of the others so it is dependent ( ie. not independent... )
• Oct 15th 2008, 05:45 PM
Hellreaver
So if the resultant isn't that matrix, the vectors aren't independent?
• Oct 15th 2008, 05:46 PM
Scopur
Quote:

Originally Posted by Hellreaver
So if the resultant isn't that matrix, the vectors aren't independent?

Yes if you don't get an identity matrix it is not linearly independent, precisely.
• Oct 15th 2008, 05:51 PM
Hellreaver
Ok, I understand. Thanks. I don't know this stuff can't just be said in plain english---> if this is this, then you have x. If not, you don't. I am having a hard time comprehending the language.
Thanks to both of you. :)
• Oct 15th 2008, 05:53 PM
Scopur
Linear algebra is the king of confusion in my mind, especially in upper years.
• Oct 15th 2008, 05:55 PM
Hellreaver
I could definitely understand, since I am confused out of my mind at lower level. Urg....
• Oct 15th 2008, 06:17 PM
ThePerfectHacker
Quote:

Originally Posted by Hellreaver
$\begin{bmatrix}1&0&0&1&0\\0&1&1/3&4/3&0\\0&0&0&0&0\\0&0&-1&-1&0\end{bmatrix}$

As Scopur said it means the equation has non-trivial solutions.

With a little more work we can actually find all the solutions. Multiply the 4th row by 1/3 and add it to the 2nd. After that multiply the 4th row by -1. Finally interchange the 3rd and 4th rows. The resulting matrix is:
$\begin{bmatrix} 1&0&0&1&0 \\ 0&1&0&1&0 \\ 0&0&1&1&0 \\ 0&0&0&0&0 \end{bmatrix}$

This tells us the following:
$1x_1+0x_2+0x_3+1x_4 = 0$
$0x_1+1x_2+0x_3+1x_4 = 0$
$0x_1+0x_2+1x_3+1x_4 = 0$
$0x_1+0x_2+0x_3+0x_4 = 0$

The fourth equation is redundant, since it is obviously true.
Solving each equation for the leading variables we get:
$x_1 = -x_4$
$x_2 = -x_4$
$x_3 = -x_4$

Let $x_4 = t$ then it means $x_1=x_2=x_3 = -x_4 = -t$.

What does this means? It means we found all solutions. Let $t$ be any number you want it to be. It follows if we set $x_1=x_2=x_3=t$ and $x_4=-t$ then we found a solution. Try it for $t=1,2,3,...$ and convince these are solutions. Since $t$ varies for infinitely many values it means there are infinitely many solutions. This is an example when the number of equations is equal to the number of variables and yet the number of solutions is infinite.