Answer
$x=\frac{1}{2}$
$y=-\frac{1}{3}$
$z=1$
Work Step by Step
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.
$z=1$
$y=-\frac{1}{3}$
$x=\frac{1}{2}$
