## Elementary Linear Algebra 7th Edition

$x=5$ $y=2$ $z=-6$
Write the augmented matrix of the system of linear equations. $\begin{bmatrix} 2 & 3 & 1 & 10\\ 2 &-3 & -3 & 22\\ 4 & -2 & 3 & -2 \end{bmatrix}$ Divide the first row by 2. $\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 2 &-3 & -3 & 22\\ 4 & -2 & 3 & -2 \end{bmatrix}$ Add -2 times the 1st row to the 2nd row to produce a new 2nd row. Add -4 times the 1st row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &-6 & -4 & 12\\ 0 & -8 & 1 & -22 \end{bmatrix}$ Divide the second row by -6. $\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &1 & \frac{2}{3} & -2\\ 0 & -8 & 1 & -22 \end{bmatrix}$ Add 8 times the 2nd row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &1 & \frac{2}{3} & -2\\ 0 & 0 & \frac{19}{3} & -38 \end{bmatrix}$ Multiply the third row by $\frac{3}{19}$ $\begin{bmatrix} 1 & \frac{3}{2} & \frac{1}{2} & 5\\ 0 &1 & \frac{2}{3} & -2\\ 0 & 0 & 1 & -6 \end{bmatrix}$ Use back-substitution to find the solution. $z=-6$ $y=2$ $x=5$