- Represent the system as an augmented matrix \([A|b]\).
- Apply row operations to achieve reduced row-echelon form.
- Eliminate variables iteratively to isolate each unknown: \(x=6, y=-9, z=7\).
- Final solution: \(\boxed{(6, -9, 7)}\).
Explanation
- Representing the System as an Augmented Matrix
To solve the system of linear equations using Gauss-Jordan elimination, we first represent the system as an augmented matrix. The system is:
\[ \begin{aligned} 2x + y - 2z &= -11 \ x + 3y - z &= -28 \ 3x + 4y - z &= -25 \end{aligned} \]
The augmented matrix \([A|b]\) is:
\[ \begin{pmatrix} 2 & 1 & -2 & | & -11 \ 1 & 3 & -1 & | & -28 \ 3 & 4 & -1 & | & -25 \end{pmatrix} \]
Our objective is to transform this matrix into reduced row-echelon form, which will give us the values of \(x, y,\) and \(z\).
- Eliminating x from Rows 2 and 3
First, we aim to get a leading 1 in the first column. We swap the first and second rows to make the calculation easier:
\[ \begin{pmatrix} 1 & 3 & -1 & | & -28 \ 2 & 1 & -2 & | & -11 \ 3 & 4 & -1 & | & -25 \end{pmatrix} \]
Now, we eliminate \(x\) from the second and third rows.
For row 2: \(R_2 = R_2 - 2 \cdot R_1\)
\(R_2 = (2, 1, -2, -11) - 2 \cdot (1, 3, -1, -28) = (0, -5, 0, 45)\)
For row 3: \(R_3 = R_3 - 3 \cdot R_1\)
\(R_3 = (3, 4, -1, -25) - 3 \cdot (1, 3, -1, -28) = (0, -5, 2, 59)\)
The matrix becomes:
\[ \begin{pmatrix} 1 & 3 & -1 & | & -28 \ 0 & -5 & 0 & | & 45 \ 0 & -5 & 2 & | & 59 \end{pmatrix} \]
- Eliminating y from Rows 1 and 3
Now we aim to get a pivot of 1 in the second row. Divide row 2 by \(-5\):
\(R_2 = R_2 / (-5) = (0, 1, 0, -9)\)
The matrix is now:
\[ \begin{pmatrix} 1 & 3 & -1 & | & -28 \ 0 & 1 & 0 & | & -9 \ 0 & -5 & 2 & | & 59 \end{pmatrix} \]
Next, eliminate the variable \(y\) from the first and third rows.
For row 1: \(R_1 = R_1 - 3 \cdot R_2\)
\(R_1 = (1, 3, -1, -28) - 3 \cdot (0, 1, 0, -9) = (1, 0, -1, -1)\)
For row 3: \(R_3 = R_3 + 5 \cdot R_2\)
\(R_3 = (0, -5, 2, 59) + 5 \cdot (0, 1, 0, -9) = (0, 0, 2, 14)\)
The matrix becomes:
\[ \begin{pmatrix} 1 & 0 & -1 & | & -1 \ 0 & 1 & 0 & | & -9 \ 0 & 0 & 2 & | & 14 \end{pmatrix} \]
- Finalizing the Reduced Row-Echelon Form
Finally, we create a pivot of 1 in the third row. Divide row 3 by 2:
\(R_3 = R_3 / 2 = (0, 0, 1, 7)\)
Now, eliminate the variable \(z\) from the first row:
For row 1: \(R_1 = R_1 + R_3\)
\(R_1 = (1, 0, -1, -1) + (0, 0, 1, 7) = (1, 0, 0, 6)\)
The reduced row-echelon form is:
\[ \begin{pmatrix} 1 & 0 & 0 & | & 6 \ 0 & 1 & 0 & | & -9 \ 0 & 0 & 1 & | & 7 \end{pmatrix} \]
Thus, the solution is \(x = 6, y = -9, z = 7\). The final result is \((x, y, z) = (\boxed{6, -9, 7})\).
Examples
Gauss-Jordan elimination is a powerful tool for solving systems of linear equations, which appear frequently in engineering, economics, and data science. For instance, imagine a company that produces three different products using three different resources (e.g., labor, steel, and energy). If the total resource consumption for different product mixes is known, Gauss-Jordan elimination allows the company to determine exactly how many units of each product were produced based on the total resources used. This ensures efficient planning and resource management, helping managers make data-driven decisions.