Question

Let \( \alpha, \beta, \) and \( \gamma \) be real numbers. Consider the following system of linear equations: 
\( x + 2y + z = 7 \)
\( x + \alpha z = 11 \)
\( 2x - 3y + \beta z = \gamma \)
Match each entry in List I to the correct entries in List II

List I		List II
(P)	If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then the system has	(1)	a unique solution
(Q)	If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then the system has	(2)	no solution
(R)	If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \), then the system has	(3)	infinitely many solutions
(S)	If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma = 28 \), then the system has	(4)	\( x = 11, y = -2 \) and \( z = 0 \) as a solution
		(5)	\( x = -15, y = 4 \) and \( z = 0 \) as a solution

• A. (P) →(3), (Q)→ (2), (R) →(1), (S) →(4)
• B. (P) →(3), (Q)→ (2), (R) →(5), (S) →(4)
• C. (P) →(2), (Q)→ (1), (R) →(4), (S) →(5)
• D. (P) →(2), (Q)→ (1), (R) →(1), (S) →(3)

Answer 1

The Correct Option is A

We have the following system of linear equations:

\(x + 2y + z = 7\) 

\(x + \alpha z = 11\)

\(2x - 3y + \beta z = \gamma\)

Let's analyze the scenarios provided in List I and match them with List II:

(P) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then the system has

When \(\beta = \frac{1}{2}(7\alpha - 3)\) and \(\gamma = 28\), the third equation becomes \((7\alpha-3)/2 = 28\). This implies the system is consistent and has a special form of \(\beta\) and \(\gamma\), leading to infinitely many solutions as it matches a dependency condition. Thus, it associates with (1) a unique solution, but due to the specific \(\beta\) form, it aligns more with (3).

(Q) If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then the system has

Here, with the condition that \(\gamma \neq 28\), the system becomes inconsistent if derived equations contradict each other, resulting in no solution. Thus, it matches with (2).

(R) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \), then the system has

With \(\alpha=1\), \(\beta\neq(7\alpha-3)/2 = 2\) and \(\gamma \neq 28\), the system is consistent and independent, leading to a unique solution. Thus it matches with (1).

(S) If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \(\gamma = 28\), then the system has

With this condition, we substitute and verify one of the solution propositions whereby, assuming a direct solution trial reveals: \(x=11\), \(y=-2\), \(z=0\) satisfies all equations. Thus, matches (4).

Hence, the correct matching is: (P) →(3), (Q)→ (2), (R) →(1), (S) →(4)

Answer 2

Solving the System of Equations 

We are given the system of equations:

\[ x + 2y + z = 7 \] \[ x + \alpha z = 11 \] \[ 2x - 3y + \beta z = \gamma \]

We can express the determinant as:

\[ \triangle = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{vmatrix} \] and from the equation, we get: \[ 3\alpha - 2(\beta - 2\alpha) - 3 = 0 \]

Solving the above equation, we find:

\[ 7\alpha - 2\beta = 3 \quad \Rightarrow \quad \beta = \frac{1}{2} (7\alpha - 3) \]

Next, we compute the determinants \( \triangle_1, \triangle_2, \) and \( \triangle_3 \):

\[ \triangle_1 = \begin{vmatrix} 7 & 2 & 1 \\ 11 & 0 & \alpha \\ \gamma & -3 & \beta \end{vmatrix}, \quad \triangle_2 = \begin{vmatrix} 1 & 7 & 1 \\ 1 & 11 & \alpha \\ 2 & \gamma & \beta \end{vmatrix}, \quad \triangle_3 = \begin{vmatrix} 1 & 7 & 2 \\ 1 & 0 & 11 \\ 2 & -3 & \gamma \end{vmatrix} \]

We solve for \( \gamma \) by simplifying \( \triangle_3 \):

\[ 33 - 2(\gamma - 22) + 7(-3) = 0 \quad \Rightarrow \quad \gamma = 28 \]

Next, we simplify \( \triangle_1 \):

\[ \triangle_1 = 21\alpha - 2(11\beta - \alpha \beta) - 33 \] \[ = 21\alpha - 22\beta + 22\beta + 2\alpha \beta - 33 \]

Next, we simplify \( \triangle_2 \):

\[ \triangle_2 = 11\beta - \alpha\beta - 7(\beta - 2\alpha) + \gamma - 22 \] \[ = 14\alpha + 4\beta + \gamma - \alpha\gamma - 22 \]

Case (P):

If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then:

\[ \triangle = 0, \quad \triangle_1 = 0, \quad \triangle_2 = 0, \quad \triangle_3 = 0 \] There are infinitely many solutions, and \( x = 11, y = -2, z = 0 \) satisfy all the three equations.

Case (Q):

If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then:

\[ \triangle = 0, \quad \text{but} \quad \triangle_3 \neq 0 \quad \text{so no solution} \]

Case (R):

If \( \beta \neq \frac{1}{2}(7\alpha - 3), \, \alpha = 1 \) and \( \gamma \neq 28 \), then:

\[ \triangle \neq 0, \quad \triangle_3 \neq 0 \quad \text{so a unique solution} \]

Case (S):

If \( \beta \neq \frac{1}{2}(7\alpha - 3), \, \alpha = 1 \) and \( \gamma = 28 \), then:

\[ \triangle \neq 0, \quad \triangle_3 = 0, \quad \triangle_1 \neq 0, \quad \triangle_2 \neq 0 \quad \text{so a unique solution} \]

Conclusion:

Thus, the correct option is (A) \( \text{(P)} \to (3), \, \text{(Q)} \to (2), \, \text{(R)} \to (1), \, \text{(S)} \to (4) \).