Answer :
To solve the given system of linear equations using Gauss-Jordan elimination, we will follow a step-by-step approach:
1. Write the augmented matrix for the system of equations:
The system of linear equations is:
[tex]\[ \begin{aligned} 2x - y + 3z &= 12 \quad \text{(Equation 1)} \\ 0x + 2y - z &= 14 \quad \text{(Equation 2)} \\ 7x - 5y + 0z &= 9 \quad \text{(Equation 3)} \end{aligned} \][/tex]
The augmented matrix is:
[tex]\[ \left[ \begin{array}{ccc|c} 2 & -1 & 3 & 12 \\ 0 & 2 & -1 & 14 \\ 7 & -5 & 0 & 9 \\ \end{array} \right] \][/tex]
2. Apply Gauss-Jordan elimination to the augmented matrix:
Step 1: Make the element in the first row, first column a 1 (pivot).
Divide the whole first row by 2:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 2 & -1 & 14 \\ 7 & -5 & 0 & 9 \\ \end{array} \right] \][/tex]
Step 2: Eliminate the first column elements for the second and third rows.
- For the third row, [tex]\( R_3 - 7R_1 \)[/tex]:
[tex]\[ R_3 = \left[ \begin{array}{ccc|c} 7 & -5 & 0 & 9 \end{array} \right] - 7\left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \end{array} \right] = \left[ \begin{array}{ccc|c} 0 & -\frac{1}{2} & -\frac{21}{2} & -33 \end{array} \right] \][/tex]
Our matrix becomes:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 2 & -1 & 14 \\ 0 & -\frac{1}{2} & -\frac{21}{2} & -33 \\ \end{array} \right] \][/tex]
Step 3: Eliminate the first column element for the third row by scaling the rows.
Multiply the second row by [tex]\( \frac{1}{2} \)[/tex] and add to the third row:
[tex]\[ R_3 = \left[ \begin{array}{ccc|c} 0 & 2 & -1 & 14 \end{array} \right] \frac{1}{2} + \left[ \begin{array}{ccc|c} 0 & -\frac{1}{2} & -\frac{21}{2} & -33 \end{array} \right] = \left[ \begin{array}{ccc|c} 0 & 0 & -11 & -26 \end{array} \right] \][/tex]
Now, the matrix becomes:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 2 & -1 & 14 \\ 0 & 0 & -11 & -26 \\ \end{array} \right] \][/tex]
Step 4: Make the element in the second row, second column a 1.
Divide the whole second row by 2:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & -11 & -26 \\ \end{array} \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & 1 & \frac{26}{11} \\ \end{array} \right] } \][/tex]
Step 5: Eliminate the second column element for the first row.
- For the first row, [tex]\( R_1 + \frac{1}{2} R_2 \)[/tex]:
[tex]\[ R_1 = \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \end{array} \right] + \frac{1}{2}\left[ \begin{array}{ccc|c} 0 & 1 & -\frac{1}{2} & 7 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 9.5 \end{array} \right] \][/tex]
Our matrix becomes:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 9.5 \quad \text{(divide row first, eliminate second and third row columns too)} \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & 1 & \frac{26}{11} \\ \end{array} \right] \][/tex]
Step 6: Make back substitution and checking final reduced rows.
\[
\begin{eqnarray}
\begin{cases}
x = 9.5\\
y = 7 \\
z = \frac{26}{11}
\end{cases}
\end{eqnarray}
}
So the final values for variables would be x = 9.5, y = 7.
1. Write the augmented matrix for the system of equations:
The system of linear equations is:
[tex]\[ \begin{aligned} 2x - y + 3z &= 12 \quad \text{(Equation 1)} \\ 0x + 2y - z &= 14 \quad \text{(Equation 2)} \\ 7x - 5y + 0z &= 9 \quad \text{(Equation 3)} \end{aligned} \][/tex]
The augmented matrix is:
[tex]\[ \left[ \begin{array}{ccc|c} 2 & -1 & 3 & 12 \\ 0 & 2 & -1 & 14 \\ 7 & -5 & 0 & 9 \\ \end{array} \right] \][/tex]
2. Apply Gauss-Jordan elimination to the augmented matrix:
Step 1: Make the element in the first row, first column a 1 (pivot).
Divide the whole first row by 2:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 2 & -1 & 14 \\ 7 & -5 & 0 & 9 \\ \end{array} \right] \][/tex]
Step 2: Eliminate the first column elements for the second and third rows.
- For the third row, [tex]\( R_3 - 7R_1 \)[/tex]:
[tex]\[ R_3 = \left[ \begin{array}{ccc|c} 7 & -5 & 0 & 9 \end{array} \right] - 7\left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \end{array} \right] = \left[ \begin{array}{ccc|c} 0 & -\frac{1}{2} & -\frac{21}{2} & -33 \end{array} \right] \][/tex]
Our matrix becomes:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 2 & -1 & 14 \\ 0 & -\frac{1}{2} & -\frac{21}{2} & -33 \\ \end{array} \right] \][/tex]
Step 3: Eliminate the first column element for the third row by scaling the rows.
Multiply the second row by [tex]\( \frac{1}{2} \)[/tex] and add to the third row:
[tex]\[ R_3 = \left[ \begin{array}{ccc|c} 0 & 2 & -1 & 14 \end{array} \right] \frac{1}{2} + \left[ \begin{array}{ccc|c} 0 & -\frac{1}{2} & -\frac{21}{2} & -33 \end{array} \right] = \left[ \begin{array}{ccc|c} 0 & 0 & -11 & -26 \end{array} \right] \][/tex]
Now, the matrix becomes:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 2 & -1 & 14 \\ 0 & 0 & -11 & -26 \\ \end{array} \right] \][/tex]
Step 4: Make the element in the second row, second column a 1.
Divide the whole second row by 2:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & -11 & -26 \\ \end{array} \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & 1 & \frac{26}{11} \\ \end{array} \right] } \][/tex]
Step 5: Eliminate the second column element for the first row.
- For the first row, [tex]\( R_1 + \frac{1}{2} R_2 \)[/tex]:
[tex]\[ R_1 = \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 6 \end{array} \right] + \frac{1}{2}\left[ \begin{array}{ccc|c} 0 & 1 & -\frac{1}{2} & 7 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 9.5 \end{array} \right] \][/tex]
Our matrix becomes:
[tex]\[ \left[ \begin{array}{ccc|c} 1 & 0 & 1 & 9.5 \quad \text{(divide row first, eliminate second and third row columns too)} \\ 0 & 1 & -\frac{1}{2} & 7 \\ 0 & 0 & 1 & \frac{26}{11} \\ \end{array} \right] \][/tex]
Step 6: Make back substitution and checking final reduced rows.
\[
\begin{eqnarray}
\begin{cases}
x = 9.5\\
y = 7 \\
z = \frac{26}{11}
\end{cases}
\end{eqnarray}
}
So the final values for variables would be x = 9.5, y = 7.
