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

$x_1 = 7x_2 - 6 x_4 + 5$ $x_2$ free $x_3 = 2x_4 + 3$ $x_4$ free
We're given the augmented matrix $$\begin{bmatrix} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ -1 & 7 & -4 & 2 & 7 \end{bmatrix}$$ First, add the first row to the third row $$\begin{bmatrix} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & -4 & 8 & 12 \end{bmatrix}$$ And add the second row to the third row $$\begin{bmatrix} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ This leaves us with the general solution to the system: $x_2$ and $x_4$ are free because they have no pivot columns. From the equations represented by the first two rows of the matrix, we have: $x_1 = 7x_2 - 6 x_4 + 5$ $x_3 = 2x_4 + 3$