1. ## Matrices

I'm solving a matrix and putting it into reduced row echelon form and the bottom ends up being 0 0 0 = 0 which I know means infinite solutions, but to finish the problem I have to write the general solution which I am not sure how to do, could someone guide me on this please? Unfortunately my professor blew right through the material and I've never seen this before. Here is the solved matrix:

1 0 0 = 7
0 1 -1 = -4
0 0 0 = 0

2. ## Re: Matrices

The general solution is: x = 7 and y = z - 4 where z is arbitrary.

Thank you!

4. ## Re: Matrices

You are welcome.

The general solution can be also written in vector form: $\begin{pmatrix}7\\ -4\\ 0\end{pmatrix}+z\begin{pmatrix}0\\ 1\\ 1\end{pmatrix}$ for arbitrary z.