System of Linear Equations Solutions for Parameters Alpha Beta Gamma

To determine the correct matching between List-I and List-II, we first analyze the given system of linear equations:

1. $x + 2y + z = 7$
2. $x + \alpha z = 11$
3. $2x - 3y + \beta z = \gamma$

We can write the coefficient matrix determinant $D$: $D = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{vmatrix}$

Expanding $D$ along the second row: $D = -1(2\beta - (-3)) - \alpha(-3 - 4) = -(2\beta + 3) + 7\alpha = 7\alpha - 2\beta - 3$

Thus, $D = 0 \iff \beta = \frac{1}{2}(7\alpha - 3)$.

Next, we reduce the augmented matrix $[A \mid B]$ using row operations:

$$\begin{bmatrix} 1 & 2 & 1 & \mid & 7 \\ 1 & 0 & \alpha & \mid & 11 \\ 2 & -3 & \beta & \mid & \gamma \end{bmatrix}$$

Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$:

$$\begin{bmatrix} 1 & 2 & 1 & \mid & 7 \\ 0 & -2 & \alpha - 1 & \mid & 4 \\ 0 & -7 & \beta - 2 & \mid & \gamma - 14 \end{bmatrix}$$

Applying $R_3 \to 2R_3 - 7R_2$:

$$\begin{bmatrix} 1 & 2 & 1 & \mid & 7 \\ 0 & -2 & \alpha - 1 & \mid & 4 \\ 0 & 0 & 2(\beta - 2) - 7(\alpha - 1) & \mid & 2(\gamma - 14) - 28 \end{bmatrix}$$

Simplifying the third row: $-(7\alpha - 2\beta - 3)z = 2(\gamma - 28)$

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Step-by-Step Analysis of List-I Entries:

• Entry (P): If $\beta = \frac{1}{2}(7\alpha - 3)$ and $\gamma = 28$:

  • Coefficient of $z$: $-(7\alpha - 2\beta - 3) = 0$
  • Constant term: $2(28 - 28) = 0$
  • The equation becomes $0 \cdot z = 0$, which gives infinitely many solutions.
  • $(P) \rightarrow (3)$

• Entry (Q): If $\beta = \frac{1}{2}(7\alpha - 3)$ and $\gamma \neq 28$:

  • Coefficient of $z$: $-(7\alpha - 2\beta - 3) = 0$
  • Constant term: $2(\gamma - 28) \neq 0$
  • The equation becomes $0 \cdot z = \text{non-zero}$, which gives no solution.
  • $(Q) \rightarrow (2)$

• Entry (R): If $\beta \neq \frac{1}{2}(7\alpha - 3)$ where $\alpha = 1$ and $\gamma \neq 28$:

  • Since $\beta \neq \frac{1}{2}(7\alpha - 3)$, we have $D \neq 0$.
  • A non-zero determinant $D$ implies that the system has a unique solution.
  • $(R) \rightarrow (1)$

• Entry (S): If $\beta \neq \frac{1}{2}(7\alpha - 3)$ where $\alpha = 1$ and $\gamma = 28$:

  • Substituting $\alpha = 1$ into equation (2): $x + (1)z = 11$.
  • Setting $z = 0$ yields $x = 11$.
  • Substituting $x = 11$ and $z = 0$ into equation (1): $11 + 2y + 0 = 7 \implies y = -2$.
  • Checking equation (3) with $x = 11, y = -2, z = 0$: $2(11) - 3(-2) + \beta(0) = 28 = \gamma$.
  • Thus, $x = 11, y = -2$ and $z = 0$ is a solution.
  • $(S) \rightarrow (4)$

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Conclusion:

The correct matching is: $(P) \rightarrow (3) \quad (Q) \rightarrow (2) \quad (R) \rightarrow (1) \quad (S) \rightarrow (4)$

This corresponds to Option A.