- Represent the system of linear equations as an augmented matrix.
- Perform elementary row operations to transform the augmented matrix into row-reduced echelon form (RREF).
- Solve for the variables by reading the identity matrix side of the final RREF matrix.
- The solution to the system is \((x, y, z) = (\boxed{1/9, -4/9, -2/3})\).
Explanation
- Analysis and Setup of the Augmented Matrix
We are given a system of three linear equations with three variables \(x, y,\) and \(z\):
- \(x + y + z = -1\)
- \(2x - y + z = 0\)
- \(x + y - 2z = 1\)
Our objective is to solve this system using Gauss-Jordan elimination, which involves transforming the augmented matrix of the system into row-reduced echelon form (RREF) by applying elementary row operations. The augmented matrix is:
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 2 & -1 & 1 & 0 \ 1 & 1 & -2 & 1 \end{array} \right] \]
- Eliminating Variables to Form an Upper Triangular Matrix
First, we eliminate the \(x\)-terms in rows 2 and 3 using row 1.
For \(R_2\): Perform \(R_2 \leftarrow R_2 - 2R_1\).
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 2-2(1) & -1-2(1) & 1-2(1) & 0-2(-1) \ 1 & 1 & -2 & 1 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 0 & -3 & -1 & 2 \ 1 & 1 & -2 & 1 \end{array} \right] \]
For \(R_3\): Perform \(R_3 \leftarrow R_3 - R_1\).
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 0 & -3 & -1 & 2 \ 1-1 & 1-1 & -2-1 & 1-(-1) \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 0 & -3 & -1 & 2 \ 0 & 0 & -3 & 2 \end{array} \right] \]
- Further Reduction of the Matrix
Next, we work on row 3 to isolate \(z\).
Perform \(R_3 \leftarrow R_3 / (-3)\):
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 0 & -3 & -1 & 2 \ 0 & 0 & 1 & -2/3 \end{array} \right] \]
Now we eliminate the \(z\)-terms in row 1 and row 2.
For \(R_2 \leftarrow R_2 + 1R_3\):
\[ 0 & -3 & -1+1 & 2 + (-2/3) = 6/3 - 2/3 = 4/3 \]
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \ 0 & -3 & 0 & 4/3 \ 0 & 0 & 1 & -2/3 \end{array} \right] \]
For \(R_1 \leftarrow R_1 - R_3\):
\[ 1 & 1 & 1-1 & -1 - (-2/3) = -3/3 + 2/3 = -1/3 \]
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 0 & -1/3 \ 0 & -3 & 0 & 4/3 \ 0 & 0 & 1 & -2/3 \end{array} \right] \]
- Final Solution Calculation
Finally, we isolate \(y\) in row 2.
Perform \(R_2 \leftarrow R_2 / (-3)\):
\[ \left[ \begin{array}{ccc|c} 1 & 1 & 0 & -1/3 \ 0 & 1 & 0 & -4/9 \ 0 & 0 & 1 & -2/3 \end{array} \right] \]
Eliminate \(y\) from row 1 by performing \(R_1 \leftarrow R_1 - R_2\):
\[ 1 & 1-1 & 0 & -1/3 - (-4/9) = -3/9 + 4/9 = 1/9 \]
The final matrix in RREF is:
\[ \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1/9 \ 0 & 1 & 0 & -4/9 \ 0 & 0 & 1 & -6/9 \end{array} \right] \]
Thus, the solution is \(x = 1/9, y = -4/9, z = -2/3\).
Examples
Gauss-Jordan elimination is widely used in economics and engineering to manage complex systems where multiple variables depend on each other. For example, in electrical engineering, it helps analyze currents in multi-loop circuits (Kirchhoff's laws). By representing the relationship between currents in different branches as a system of linear equations, engineers can use matrix reduction to determine the exact current flowing through every component in the circuit, ensuring that the design functions safely and efficiently.