- Simplify the given system and verify consistency using the augmented matrix: \(x - y = -1\) and \(z = 0\).
- Recognize that the third equation is a dependent linear combination of the first two, resulting in a row of zeros in the row-echelon form.
- Identify the presence of a free variable (\(y=t\)), which implies infinitely many solutions for the system.
- Final answer: The system has \(\infty\) solutions, expressed as \((t-1, t, 0)\) for any \(t \in \mathbb{R}\).
- Final Answer: \(\boxed{\text{Yes}}\)
Explanation
- Analyzing the System of Equations
We are given a system of three linear equations with three variables:
\[ \begin{cases} x - y + 2z = -1 \ -3x + 3y + 5z = 3 \ 2x - 2y = -2 \end{cases} \]
To analyze the nature of the solutions, we look at the system's structure. We observe that the third equation, \(2x - 2y = -2\), can be simplified by dividing by 2 to get \(x - y = -1\). This implies that \(x = y - 1\). If we substitute \(x - y = -1\) into the first equation, we get \(-1 + 2z = -1\), which simplifies to \(2z = 0\), or \(z = 0\). Finally, we check the consistency by substituting \(x - y = -1\) and \(z = 0\) into the second equation: \(-3(x - y) + 5z = -3(-1) + 5(0) = 3 + 0 = 3\), which holds true. Thus, the entire system is equivalent to the conditions \(z = 0\) and \(x - y = -1\).
- Examining the Augmented Matrix
The provided augmented matrix for the system is:
\[ \left[\begin{array}{rrrr} 1 & -1 & 0 & -1 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 \end{array}\right] \]
This matrix corresponds to the system:
\[ \begin{cases} 1x - 1y + 0z = -1 \ 0x + 0y + 1z = 0 \ 0x + 0y + 0z = 0 \end{cases} \]
This confirms our earlier deduction that \(z = 0\) and \(x - y = -1\). The row of zeros in the bottom row indicates that the third equation is a linear combination of the other two (it is dependent). Since we have 3 variables and only 2 independent equations, the system is underdetermined.
- Conclusion on the Nature of the Solutions
Because the system has fewer independent equations than variables, we have a free variable. We can express \(x\) in terms of \(y\): let \(y = t\), where \(t\) is any real number. Then \(x = t - 1\) and \(z = 0\). Since \(t\) can be any real number, there are infinitely many solutions of the form \((t-1, t, 0)\). Therefore, the system indeed has infinitely many solutions.
Examples
This problem demonstrates how to solve linear systems that have infinitely many solutions, a concept widely used in fields like computer graphics and structural engineering. For instance, in structural analysis, determining the forces within a truss system often involves solving sets of linear equations; if the structure is statically indeterminate, it may result in infinitely many solutions representing various states of equilibrium the structure can maintain. By identifying these dependencies, engineers can optimize designs and ensure stability.