The set of solutions is $\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}$

Step-by-step explanation:

The augmented matrix of the system is $\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]$ .

We will use rows operations for find the echelon form of the matrix.

In row 2 we subtract $\frac{2}{3}$ from row 1. (R2- 2/3R1) and we obtain the matrix $\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]$
We multiply the row 1 by $\frac{1}{3}$ .

Now we solve for the unknown variables:

$x_2+x_3=-7$ , $x_2=-7-x_3$
$x_1+2x_2+2x_3=-2$ , $x_1+2(-7-x_3)+2x_3=-2$ then $x_1=12$

The system has a free variable, the the system has infinite solutions and the set of solutions is $\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}$

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3 years ago