Did I solve the system of equations correctly and if yes, how would the set of solution look like?

Question

I wanted to ask whether I solved the following system of equations correctly:

Solve the following (augmented matrix, right vector is the solution vector) $$ \left[\begin{array}{rrrr|r} -4 & -8 & 8 & -8 & 12 \\ 0 & -6 & 0 & -12 & 6\\ -6 & -6 & 12 & 0 & 12 \end{array}\right] $$

I solved it this far:

$$ \left[\begin{array}{rrrr|r} 1 & 1 & -2 & 0 & -2 \\ 0 & 1 & 0 & 2 & -1\\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$

Is it correct and if yes how would I express the set of solutions?

Answer 1

No, it is not correct. If you follow the standard procedure, what you should get is$$\left[\begin{array}{rrrr|r}1 & 0 & -2 & 0 & -2 \\ 0 & 1 & 0 & 2 & -1\\ 0 & 0 & 0 & 1 & -\frac12 \end{array}\right].$$Therefore, the set of solutions is$$\left\{\left(x,0,1+\frac x2,-\frac12\right)\,\middle|\,x\in\Bbb R\right\}.$$

Answer 2

You need one more step to zero the $(1,2)$-entry. The general solution will have $x_3$ and $x_4$ as free variables.

To check the preliminary work that you showed, write each of your original rows as a linear combination of the first two rows of the reduced form.