Solve the following system of linear equations: 3x1+6x2+6x3 = -9 -2x1–3x2-3x3 = 3 If the system has infinitely many solutions, your answer may use expressions involving the parameters r, s, and t. O The system has at least one solution x1 = 0 x2 = 0 X3 = 0 O O

The set of solutions is [tex]\{\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} \}[/tex]

Step-by-step explanation:

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

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

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

Now we solve for the unknown variables:

[tex]x_2+x_3=-7[/tex], [tex]x_2=-7-x_3[/tex]
[tex]x_1+2x_2+2x_3=-2[/tex], [tex]x_1+2(-7-x_3)+2x_3=-2[/tex] then [tex]x_1=12[/tex]

The system has a free variable, the the system has infinite solutions and the set of solutions is [tex]\{\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} \}[/tex]