Basic Augmented Matrix Problem

[kamui8899]

I have an augmented matrix:

1 2 3 4
0 3 4 5
3 12 1 2

Now this matrix simplifies to:

1 2 3 4
0 3 4 5
0 0 0 0

So there are infinite solutions, however I have to write all the solutions to the equation. I wanted to do this with vectors.

So I first solved for x and for y and got:

y = (4/3)z - 5/3

x = -2((4/3)z - 5/3) -3z + 4

So now what do I do, can I simply write the answer as:

<-2((4/3)z - 5/3) -3z + 4, (4/3)z -5/3, z>

which then goes to

z<-2(4/3) - 3, 4/3, 1> + <-2(-5/3) +4, 5/3, 0>

?

This doesn't seem correct to me though. Did I solve the equation correctly, and if so, is what I wrote the correct answer for finding all solutions to the equation?

Thanks for the help.

 

[Galileo]

You did the reduction wrong. The bottom row does not become 0.