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Agata [3.3K]
3 years ago
5

Solve the following system of linear equations: 3x1+6x2+6x3 = -9 -2x1–3x2-3x3 = 3 If the system has infinitely many solutions, y

our 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
Mathematics
1 answer:
choli [55]3 years ago
6 0

Answer:

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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