- Represent the system of equations as an augmented matrix: \(\left[\begin{array}{ccc|c}3 & -1 & 2 & 5 \ 1 & -1 & 2 & 1 \ 5 & -2 & 4 & 19 \end{array}\right]\).
- Apply elementary row operations to reduce the matrix to row-echelon form.
- Observe that the third row leads to a contradiction, \(0 = 11\), implying the system is inconsistent.
- Conclude that there is no solution: \(\boxed{\text{No solution}}\).
Explanation
- Problem Analysis and Augmented Matrix Representation
We are given the following system of linear equations:
\[ \begin{array}{r@{\quad}l} 3x - y + 2z = 5 & (1) \ x - y + 2z = 1 & (2) \ 5x - 2y + 4z = 19 & (3) \end{array} \]
Our objective is to solve this system using the Gauss-Jordan elimination method. We start by representing the system as an augmented matrix \([A|b]\):
\[ \left[\begin{array}{ccc|c} 3 & -1 & 2 & 5 \ 1 & -1 & 2 & 1 \ 5 & -2 & 4 & 19 \end{array}\right] \]
- Row Reduction Operations
To simplify the matrix, we perform row operations. First, swap Row 1 and Row 2 to obtain a pivot of 1 in the first position:
\[ \left[\begin{array}{ccc|c} 1 & -1 & 2 & 1 \ 3 & -1 & 2 & 5 \ 5 & -2 & 4 & 19 \end{array}\right] \]
Next, we eliminate the \(x\)-terms in the second and third rows by performing row operations:
- \(R_2 \leftarrow R_2 - 3R_1\)
- \(R_3 \leftarrow R_3 - 5R_1\)
Calculating these operations:
For \(R_2\): \((3 - 3 \times 1, -1 - 3 \times (-1), 2 - 3 \times 2, 5 - 3 \times 1) = (0, 2, -4, 2)\).
For \(R_3\): \((5 - 5 \times 1, -2 - 5 \times (-1), 4 - 5 \times 2, 19 - 5 \times 1) = (0, 3, -6, 14)\).
The matrix becomes:
\[ \left[\begin{array}{ccc|c} 1 & -1 & 2 & 1 \ 0 & 2 & -4 & 2 \ 0 & 3 & -6 & 14 \end{array}\right] \]
- Conclusion and Final Answer
Next, simplify the second row by dividing it by 2:
\(R_2 \leftarrow \frac{1}{2} R_2\) gives \((0, 1, -2, 1)\).
The matrix is now:
\[ \left[\begin{array}{ccc|c} 1 & -1 & 2 & 1 \ 0 & 1 & -2 & 1 \ 0 & 3 & -6 & 14 \end{array}\right] \]
Now, eliminate the \(y\)-term in the third row:
\(R_3 \leftarrow R_3 - 3R_2\)
Calculation: \((0 - 3 \times 0, 3 - 3 \times 1, -6 - 3 \times (-2), 14 - 3 \times 1) = (0, 0, 0, 11)\).
The resulting matrix is:
\[ \left[\begin{array}{ccc|c} 1 & -1 & 2 & 1 \ 0 & 1 & -2 & 1 \ 0 & 0 & 0 & 11 \end{array}\right] \]
The third row corresponds to the equation \(0x + 0y + 0z = 11\), which is \(0 = 11\). This is a contradiction. Therefore, the system is inconsistent and has no solutions.
Examples
Solving systems of linear equations is fundamental in fields such as engineering, economics, and data science. For instance, in structural engineering, when analyzing the forces and tensions on different parts of a bridge, one sets up a system of linear equations based on the equilibrium of forces. If the equations are contradictory, as seen in this example, it might indicate that the model is physically impossible (e.g., the structure is unstable or cannot exist under the given constraints), helping engineers identify errors in their design or assumptions.