ANSWER:
x =- 4
y = 3
z = 2
STEP-BY-STEP EXPLANATION:
We have the following system of equations:
[tex]\begin{gathered} 3x-2y-3z=-24 \\ \\ 3x+5y+2z=7 \\ \\ -x+5y+3z=25 \end{gathered}[/tex]The first thing is to build the matrix from the system of equations, just like this:
[tex]\begin{bmatrix}3&-2&-3&¦&-24\\ 3&5&2&¦&7\\ -1&5&3&¦&25\end{bmatrix}[/tex]We solve by the Gauss-Jordan method, thus:
[tex]\begin{gathered} \text{ We find the pivot in column number 1 \lparen reversing the sign in the whole row\rparen and change row number 3 to number 1} \\ \\ \begin{bmatrix}1 & -5 & -3 & ¦ & -25 \\ 3 & 5 & 2 & ¦ & 7 \\ 3 & -2 & -3 & ¦ & -24\end{bmatrix} \\ \\ \text{ We multiply row number 1 by 3} \\ \\ \begin{bmatrix}3 & -15 & -9 & ¦ & -75 \\ 3 & 5 & 2 & ¦ & 7 \\ 3 & -2 & -3 & ¦ & -24\end{bmatrix} \\ \\ \text{ We subtract row number 1 by row number 2} \\ \\ \begin{bmatrix}3 & -15 & -9 & ¦ & -75 \\ 0 & 20 & 11 & ¦ & 82 \\ 3 & -2 & -3 & ¦ & -24\end{bmatrix} \\ \\ \text{ We subtract row number 1 from row number 3 and restore it} \\ \\ \begin{bmatrix}1 & -5 & -3 & ¦ & -25 \\ 0 & 20 & 11 & ¦ & 82 \\ 0 & 13 & 6 & ¦ & 51\end{bmatrix} \\ \\ \text{ We find the pivot in column number 2 by dividing row number 2 by 20} \\ \\ \begin{bmatrix}1 & -5 & -3 & ¦ & -25 \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 13 & 6 & ¦ & 51\end{bmatrix} \\ \\ \text{ We multiply row number 2 by -5} \\ \\ \begin{bmatrix}1 & -5 & -3 & ¦ & -25 \\ 0 & -5 & -\frac{11}{4} & ¦ & -\frac{41}{2} \\ 0 & 13 & 6 & ¦ & 51\end{bmatrix} \\ \\ \text{ We subtract row number 2 from row number 1 and restore it} \\ \\ \begin{bmatrix}1 & 0 & -\frac{1}{4} & ¦ & -\frac{9}{2} \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 13 & 6 & ¦ & 51\end{bmatrix} \\ \\ \text{ We multiply row number 2 by 13} \\ \\ \begin{bmatrix}1 & 0 & -\frac{1}{4} & ¦ & -\frac{9}{2} \\ 0 & 13 & \frac{143}{20} & ¦ & \frac{533}{10} \\ 0 & 13 & 6 & ¦ & 51\end{bmatrix} \\ \\ \text{ We subtract row number 2 from row number 3 and restore it} \\ \\ \begin{bmatrix}1 & 0 & -\frac{1}{4} & ¦ & -\frac{9}{2} \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 0 & -\frac{23}{20} & ¦ & -\frac{23}{10}\end{bmatrix} \\ \\ \text{ We find the pivot in column number 3 by dividing row number 3 by -23/20} \\ \\ \begin{bmatrix}1 & 0 & -\frac{1}{4} & ¦ & -\frac{9}{2} \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 0 & 1 & ¦ & 2\end{bmatrix} \\ \\ \text{ We multiply row number 3 by -1/4} \\ \\ \begin{bmatrix}1 & 0 & -\frac{1}{4} & ¦ & -\frac{9}{2} \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 0 & -\frac{1}{4} & ¦ & -\frac{1}{2}\end{bmatrix} \\ \\ \text{ We subtract row number 3 from row number 1 and restore it} \\ \\ \begin{bmatrix}1 & 0 & 0 & ¦ & -4 \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 0 & 1 & ¦ & 2\end{bmatrix} \\ \\ \text{ We multiply row number 3 by 11/20} \\ \\ \begin{bmatrix}1 & 0 & 0 & ¦ & -4 \\ 0 & 1 & \frac{11}{20} & ¦ & \frac{82}{20} \\ 0 & 0 & \frac{11}{20} & ¦ & \frac{11}{10}\end{bmatrix} \\ \\ \text{ We subtract row number 3 from row number 2 and restore it} \\ \\ \begin{bmatrix}1 & 0 & 0 & ¦ & -4 \\ 0 & 1 & 0 & ¦ & 3 \\ 0 & 0 & 1 & ¦ & 2\end{bmatrix} \\ \\ \text{ Therefore:} \\ \\ x=-4 \\ \\ y=3 \\ \\ z=2 \end{gathered}[/tex]Therefore, the solution of the system is x =- 4, y = 3 and z = 2
