Write the solutions that can be read from the matrix.

[tex]\[
\left[\begin{array}{rrrrrrr}
x_1 & x_2 & x_3 & s_1 & s_2 & z & \\
3 & 1 & 2 & 3 & 0 & 0 & 17 \\
2 & 0 & 1 & 0 & 1 & 0 & 14 \\
\hline -2 & 0 & 0 & 1 & 0 & 1 & 15
\end{array}\right]
\][/tex]

Type the values for the solutions:

[tex]\[
\begin{array}{l}
x_1 = \square \quad x_2 = \square \quad x_3 = \square \\
s_1 = \square \quad s_2 = \square \quad z = \square
\end{array}
\][/tex]



Answer :

To determine the solutions from the matrix provided, we interpret the coefficients and the final column values in each row as follows:

[tex]\[ \left[\begin{array}{rrrrrrr} x_1 & x_2 & x_3 & s_1 & s_2 & z & \\ 3 & 1 & 2 & 3 & 0 & 0 & 17 \\ 2 & 0 & 1 & 0 & 1 & 0 & 14 \\ \hline -2 & 0 & 0 & 1 & 0 & 1 & 15 \end{array}\right] \][/tex]

From this matrix:
1. The first row equation is:
[tex]\[ 3x_1 + 1x_2 + 2x_3 + 3s_1 = 17 \][/tex]
Solving for [tex]\(x_1\)[/tex], set [tex]\(x_2\)[/tex], [tex]\(x_3\)[/tex], and [tex]\(s_1\)[/tex] to 0 and solve for [tex]\(x_1\)[/tex]:
[tex]\[ 3x_1 = 17 \implies x_1 = \frac{17}{3} \approx 5.666666666666667 \][/tex]

2. The second row equation is:
[tex]\[ 2x_1 + 1s_2 = 14 \][/tex]
Solving for [tex]\(s_2\)[/tex], set [tex]\(x_1\)[/tex] to 0:
[tex]\[ s_2 = 14 \][/tex]

3. The third row is the objective function:
[tex]\[ -2x_1 + 1s_1 + 1z = 15 \][/tex]
Solving for [tex]\(z\)[/tex] when [tex]\(x_1 = 0\)[/tex] and [tex]\(s_1 = 0\)[/tex]:
[tex]\[ z = 15 \][/tex]

From the matrix, we can read off the values as:
[tex]\[ \begin{array}{l} x_1 = 5.666666666666667 \quad x_2 = 0 \quad x_3 = 0 \\ s_1 = 0 \quad s_2 = 14 \quad z = 15 \end{array} \][/tex]

Therefore, the values of the solutions are:
[tex]\[ \begin{array}{ll} x_1 = 5.666666666666667 \quad x_2 = 0 \quad x_3 = 0 \\ s_1 = 0 \quad s_2 = 14 \quad z = 15 \end{array} \][/tex]