Back-substitution help

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January 26th 2011, 07:18 AM
learningguy

Back-substitution help

The answer to this system is probably really simple but the book doesn't explain it very well, so I'm asking you guys.

x+y+z=0
y=0

I'm not sure if the answer is no solution or if there is a way I can go further by maybe doing something like x+z=0-y. Any help would be much appreciated.
January 26th 2011, 07:24 AM
TheEmptySet

Quote:

Originally Posted by learningguy

The answer to this system is probably really simple but the book doesn't explain it very well, so I'm asking you guys.

x+y+z=0
y=0

I'm not sure if the answer is no solution or if there is a way I can go further by maybe doing something like x+z=0-y. Any help would be much appreciated.

What you have is an under determined system.

If you write this as an augmented matrix you get

$\begin{bmatrix} 1 & 1 & 1 &0 \\ 0 & 1& 0 & 0 \end{bmatrix}$

Since the z column does not have a leading 1 it is you free parameter.
So lets set $z=t, \, \, t \in \mathbb{R}$
We know that $y=0$ and that $x=-z=-t$

So this gives the solution to the system as

$\begin{pmatrix} -t \\ 0 \\t\end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\1\end{pmatrix}t$

So every solution to this system is one the above line.
January 26th 2011, 07:44 AM
learningguy

Thanks alot, sometimes I think I make stuff too hard after you explained I now see how the book explained it after looking through some of the examples again but it didn't make it very obvious. I have another small question too though. There is another question that wants me to solve a system.

1/2x - 1/3y = 1
-2x + 4/3y = -4

In order to solve that system I think I would have to multiply the top equation by 4 but then that would make the answer 0 = 0. Would the answer to that system be no solution?