Answer :
Sure! Let's solve the system of equations using the Gauss-Jordan elimination method. The given system is:
1. [tex]\(3x_1 + 6x_2 - 16x_3 = -14\)[/tex]
2. [tex]\(2x_1 + 19x_2 - 50x_3 = -13\)[/tex]
3. [tex]\(x_1 + 5x_2 - 13x_3 = -5\)[/tex]
Step 1: Write the augmented matrix for the system.
[tex]\[
\begin{bmatrix}
3 & 6 & -16 & | & -14 \\
2 & 19 & -50 & | & -13 \\
1 & 5 & -13 & | & -5
\end{bmatrix}
\][/tex]
Step 2: Start with the first row and make the pivot (first element in the first row) equal to 1.
Divide the entire first row by 3:
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
2 & 19 & -50 & | & -13 \\
1 & 5 & -13 & | & -5
\end{bmatrix}
\][/tex]
Step 3: Use the first row to make the first element of the second and third rows equal to 0.
For the second row: Subtract 2 times the first row from the second row.
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
0 & 15 & -\frac{82}{3} & | & \frac{15}{3} \\
1 & 5 & -13 & | & -5
\end{bmatrix}
\][/tex]
For the third row: Subtract the first row from the third row.
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
0 & 15 & -\frac{82}{3} & | & \frac{15}{3} \\
0 & 3 & -\frac{23}{3} & | & \frac{1}{3}
\end{bmatrix}
\][/tex]
Step 4: Make the pivot of the second row equal to 1.
Divide the entire second row by 15:
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 3 & -\frac{23}{3} & | & \frac{1}{3}
\end{bmatrix}
\][/tex]
Step 5: Use the second row to make the second elements of the first and third rows equal to 0.
For the first row: Subtract 2 times the second row from the first row.
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{110}{45} & | & -\frac{5}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 3 & -\frac{23}{3} & | & \frac{1}{3}
\end{bmatrix}
\][/tex]
For the third row: Subtract 3 times the second row from the third row.
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{110}{45} & | & -\frac{5}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 0 & -\frac{1}{3} & | & -\frac{22}{9}
\end{bmatrix}
\][/tex]
Step 6: Make the pivot of the third row equal to 1.
Multiply the third row by -3:
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{110}{45} & | & -\frac{5}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 0 & 1 & | & 2
\end{bmatrix}
\][/tex]
Step 7: Use the third row to make the third elements of the first and second rows equal to 0.
For the first row: Subtract [tex]\(\frac{110}{45}\)[/tex] times the third row from the first row.
[tex]\[
\begin{bmatrix}
1 & 0 & 0 & | & -4 \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 0 & 1 & | & 2
\end{bmatrix}
\][/tex]
For the second row: Add [tex]\(\frac{82}{45}\)[/tex] times the third row to the second row.
[tex]\[
\begin{bmatrix}
1 & 0 & 0 & | & -4 \\
0 & 1 & 0 & | & 5 \\
0 & 0 & 1 & | & 2
\end{bmatrix}
\][/tex]
This gives us the solutions:
[tex]\(x_1 = -4\)[/tex], [tex]\(x_2 = 5\)[/tex], [tex]\(x_3 = 2\)[/tex].
So, the solution to the system of equations is:
- [tex]\(x_1 = -4\)[/tex]
- [tex]\(x_2 = 5\)[/tex]
- [tex]\(x_3 = 2\)[/tex]
1. [tex]\(3x_1 + 6x_2 - 16x_3 = -14\)[/tex]
2. [tex]\(2x_1 + 19x_2 - 50x_3 = -13\)[/tex]
3. [tex]\(x_1 + 5x_2 - 13x_3 = -5\)[/tex]
Step 1: Write the augmented matrix for the system.
[tex]\[
\begin{bmatrix}
3 & 6 & -16 & | & -14 \\
2 & 19 & -50 & | & -13 \\
1 & 5 & -13 & | & -5
\end{bmatrix}
\][/tex]
Step 2: Start with the first row and make the pivot (first element in the first row) equal to 1.
Divide the entire first row by 3:
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
2 & 19 & -50 & | & -13 \\
1 & 5 & -13 & | & -5
\end{bmatrix}
\][/tex]
Step 3: Use the first row to make the first element of the second and third rows equal to 0.
For the second row: Subtract 2 times the first row from the second row.
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
0 & 15 & -\frac{82}{3} & | & \frac{15}{3} \\
1 & 5 & -13 & | & -5
\end{bmatrix}
\][/tex]
For the third row: Subtract the first row from the third row.
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
0 & 15 & -\frac{82}{3} & | & \frac{15}{3} \\
0 & 3 & -\frac{23}{3} & | & \frac{1}{3}
\end{bmatrix}
\][/tex]
Step 4: Make the pivot of the second row equal to 1.
Divide the entire second row by 15:
[tex]\[
\begin{bmatrix}
1 & 2 & -\frac{16}{3} & | & -\frac{14}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 3 & -\frac{23}{3} & | & \frac{1}{3}
\end{bmatrix}
\][/tex]
Step 5: Use the second row to make the second elements of the first and third rows equal to 0.
For the first row: Subtract 2 times the second row from the first row.
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{110}{45} & | & -\frac{5}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 3 & -\frac{23}{3} & | & \frac{1}{3}
\end{bmatrix}
\][/tex]
For the third row: Subtract 3 times the second row from the third row.
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{110}{45} & | & -\frac{5}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 0 & -\frac{1}{3} & | & -\frac{22}{9}
\end{bmatrix}
\][/tex]
Step 6: Make the pivot of the third row equal to 1.
Multiply the third row by -3:
[tex]\[
\begin{bmatrix}
1 & 0 & \frac{110}{45} & | & -\frac{5}{3} \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 0 & 1 & | & 2
\end{bmatrix}
\][/tex]
Step 7: Use the third row to make the third elements of the first and second rows equal to 0.
For the first row: Subtract [tex]\(\frac{110}{45}\)[/tex] times the third row from the first row.
[tex]\[
\begin{bmatrix}
1 & 0 & 0 & | & -4 \\
0 & 1 & -\frac{82}{45} & | & \frac{1}{3} \\
0 & 0 & 1 & | & 2
\end{bmatrix}
\][/tex]
For the second row: Add [tex]\(\frac{82}{45}\)[/tex] times the third row to the second row.
[tex]\[
\begin{bmatrix}
1 & 0 & 0 & | & -4 \\
0 & 1 & 0 & | & 5 \\
0 & 0 & 1 & | & 2
\end{bmatrix}
\][/tex]
This gives us the solutions:
[tex]\(x_1 = -4\)[/tex], [tex]\(x_2 = 5\)[/tex], [tex]\(x_3 = 2\)[/tex].
So, the solution to the system of equations is:
- [tex]\(x_1 = -4\)[/tex]
- [tex]\(x_2 = 5\)[/tex]
- [tex]\(x_3 = 2\)[/tex]
