## Elementary Linear Algebra 7th Edition

$x=\frac{1}{2}$ $y=-\frac{1}{3}$ $z=1$
Write the augmented matrix of the system of linear equations. $\begin{bmatrix} 2 & 3 & 3 & 3\\ 6 & 6 & 12 &13\\ 12 & 9 &-1 & 2 \end{bmatrix}$ Interchange the first and the second row. $\begin{bmatrix} 6 & 6 & 12 &13\\ 2 & 3 & 3 & 3\\ 12 & 9 &-1 & 2\\ \end{bmatrix}$ Divide the first row by 6. $\begin{bmatrix} 1 & 1 & 2 &\frac{13}{6}\\ 2 & 3 & 3 & 3\\ 12 & 9 &-1 & 2\\ \end{bmatrix}$ Add -2 times the 1st row to the 2nd row to produce a new 2nd row. Add -12 times the 1st row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & 1 & 2 &\frac{13}{6}\\ 0 & 1& -1 & -\frac{4}{3}\\ 0 & -3 &-25 &-24\\ \end{bmatrix}$ Add 3 times the 2nd row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & 1 & 2 &\frac{13}{6}\\ 0 & 1& -1 & -\frac{4}{3}\\ 0 & 0 &-28 &-28\\ \end{bmatrix}$ Divide the third row by -28. $\begin{bmatrix} 1 & 1 & 2 &\frac{13}{6}\\ 0 & 1& -1 & -\frac{4}{3}\\ 0 & 0 &1 &1\\ \end{bmatrix}$ Use back-substitution to find the solution. $z=1$ $y=-\frac{1}{3}$ $x=\frac{1}{2}$