Linear Algebra and Its Applications (5th Edition)

$x_3$ and $x_4$ free $$x_1=-7x_3-9$$ $$x_2=6x_3+3x_4+2$$ $$x_4=0$$
We're given the augmented matrix: $$\begin{bmatrix} 1 & 2 & -5 & -6 & 0 & -5 \\ 0 & 1 & -6 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ To put it into reduced row echelon form, add $-2$ times the second row to the first row: $$\begin{bmatrix} 1 & 0 & 7 & 0 & 0 & -9 \\ 0 & 1 & -6 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ Now, we can see that $x_3$ and $x_4$ are free. From the matrix, we have: $$x_1=-7x_3-9$$ $$x_2=6x_3+3x_4+2$$ $$x_4=0$$