## Linear Algebra and Its Applications (5th Edition)

$x_3$ and $x_5$ are free $$x_1=3x_5+5$$ $$x_2=4x_5+1$$ $$x_4=-9x_5+4$$
We're given the augmented matrix $$\begin{bmatrix} 1 & -3 & 0 & -1 & 0 & -2 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ First, add $3$ times the second row to the first $$\begin{bmatrix} 1 & 0 & 0 & -1 & -12 & 1 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ And add the third row to the first. $$\begin{bmatrix} 1 & 0 & 0 & 0 & -3 & 5 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ The matrix is now in reduced row echelon form, so we can see the solution to the system: $x_3$ and $x_5$ are free because their corresponding columns have no pivot entries. From the first three rows, we can clearly see that $$x_1=3x_5+5$$ $$x_2=4x_5+1$$ $$x_4=-9x_5+4$$