Problem Statement
We are given a system of linear equations with real numbers \(\alpha\), \(\beta\), and \(\gamma\):
\[
\begin{cases}
x + 2y + z = 7 \\
x + \alpha z = 11 \\
2x – 3y + \beta z = \gamma
\end{cases}
\]
Our goal is to analyze this system under different conditions and match each scenario to its corresponding solution type.
Lists to Match
List-I: Scenarios
- (P) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma = 28\), then the system has:
- (Q) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma \neq 28\), then the system has:
- (R) If \(\beta \neq \frac{1}{2}(7\alpha – 3)\) where \(\alpha = 1\) and \(\gamma \neq 28\), then the system has:
- (S) If \(\beta \neq \frac{1}{2}(7\alpha – 3)\) where \(\alpha = 1\) and \(\gamma = 28\), then the system has:
List-II: Solutions
- A unique solution
- No solution
- Infinitely many solutions
- \(x = 11\), \(y = -2\), and \(z = 0\) as a solution
- \(x = -15\), \(y = 4\), and \(z = 0\) as a solution
Options
- (A) (P) \(\rightarrow\) (3), (Q) \(\rightarrow\) (2), (R) \(\rightarrow\) (1), (S) \(\rightarrow\) (4)
- (B) (P) \(\rightarrow\) (3), (Q) \(\rightarrow\) (2), (R) \(\rightarrow\) (5), (S) \(\rightarrow\) (4)
- (C) (P) \(\rightarrow\) (2), (Q) \(\rightarrow\) (1), (R) \(\rightarrow\) (4), (S) \(\rightarrow\) (5)
- (D) (P) \(\rightarrow\) (2), (Q) \(\rightarrow\) (1), (R) \(\rightarrow\) (1), (S) \(\rightarrow\) (3)
Step-by-Step Solution
Step 1: Analyze the System
The system of equations is:
\[
\begin{cases}
x + 2y + z = 7 \quad \text{(Equation 1)} \\
x + \alpha z = 11 \quad \text{(Equation 2)} \\
2x – 3y + \beta z = \gamma \quad \text{(Equation 3)}
\end{cases}
\]
We need to determine the nature of the solutions based on the values of \(\alpha\), \(\beta\), and \(\gamma\).
Step 2: Match Scenarios to Solutions
Let’s analyze each scenario in List-I:
- (P) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma = 28\):
- The system has infinitely many solutions (Option 3).
- (Q) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma \neq 28\):
- The system has no solution (Option 2).
- (R) If \(\beta \neq \frac{1}{2}(7\alpha – 3)\) where \(\alpha = 1\) and \(\gamma \neq 28\):
- The system has a unique solution (Option 1).
- (S) If \(\beta \neq \frac{1}{2}(7\alpha – 3)\) where \(\alpha = 1\) and \(\gamma = 28\):
- The system has \(x = 11\), \(y = -2\), and \(z = 0\) as a solution (Option 4).
Step 3: Verify the Correct Option
Based on the above analysis, the correct matching is:
- (P) \(\rightarrow\) (3)
- (Q) \(\rightarrow\) (2)
- (R) \(\rightarrow\) (1)
- (S) \(\rightarrow\) (4)
Thus, the correct option is (A).
Conclusion
This problem demonstrates how parameters like \(\alpha\), \(\beta\), and \(\gamma\) can influence the solutions of a system of linear equations. By carefully analyzing the conditions, we can determine whether the system has a unique solution, no solution, or infinitely many solutions.