One variable equal to 0 in linear equation system.

Question

I have a hard time getting a hang of a certain concept when it comes to linear equation systems.

Say that I have the equation system translated to a matrice:

\begin{matrix} 2 & -2 & 3 & 0 \\ 0 & 2 & -2 & 0 \\ 1 & 0 & 3 & 0 \\ \end{matrix}

Eventually what I will get is:

\begin{matrix} 1 & -1 & \frac 32 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & \frac 52 & 0 \\ \end{matrix}

\begin{matrix} 1 & -1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ \end{matrix} $$ \left\{ \begin{array}{c} x-y+z=0 _1 \\ y-z= 0 \\ z=0 \end{array} \right. $$

And I was wondering if someone could explain how you're supposed to interpret the answer. Is z a free variable here?

Answer 1

$z$ is not a free variable. If you were to continue your row reductions into RREF form, you obtain:

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

which shows that $x = y = z = 0$, i.e. you obtain a unique solution to the system.