From tech troubles to travel tips, IDNLearn.com has answers to all your questions. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
Sagot :
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.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.
