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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

  1. (P) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma = 28\), then the system has:
  2. (Q) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma \neq 28\), then the system has:
  3. (R) If \(\beta \neq \frac{1}{2}(7\alpha – 3)\) where \(\alpha = 1\) and \(\gamma \neq 28\), then the system has:
  4. (S) If \(\beta \neq \frac{1}{2}(7\alpha – 3)\) where \(\alpha = 1\) and \(\gamma = 28\), then the system has:

List-II: Solutions

  1. A unique solution
  2. No solution
  3. Infinitely many solutions
  4. \(x = 11\), \(y = -2\), and \(z = 0\) as a solution
  5. \(x = -15\), \(y = 4\), 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:

  1. (P) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma = 28\):
    • The system has infinitely many solutions (Option 3).
  2. (Q) If \(\beta = \frac{1}{2}(7\alpha – 3)\) and \(\gamma \neq 28\):
    • The system has no solution (Option 2).
  3. (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).
  4. (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.

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