JEE Advanced 2023 Paper 1 · Q14 · Systems of Equations

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:
(P) If $\beta = \dfrac{1}{2}(7\alpha - 3)$ and $\gamma = 28$, then the system has
(Q) If $\beta = \dfrac{1}{2}(7\alpha - 3)$ and $\gamma 
e 28$, then the system has
(R) If $\beta 
e \dfrac{1}{2}(7\alpha - 3)$ where $\alpha = 1$ and $\gamma 
e 28$, then the system has
(S) If $\beta 
e \dfrac{1}{2}(7\alpha - 3)$ where $\alpha = 1$ and $\gamma = 28$, then the system has
List-II:
(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

SolveKar AI · Accepted Answer

Determinant $\Delta = 7 - 2\beta - 3\alpha$. Setting $\Delta = 0$ gives $\beta = (7\alpha - 3)/2$. Compute $\Delta_z$ and check consistency. (P) $\Delta=0$, $\gamma=28$: $\Delta_x = \Delta_y = \Delta_z = 0$, infinitely many — (3). (Q) $\Delta=0$, $\gamma
e 28$: $\Delta_z 
e 0$, no solution — (2). (R) $\Delta
e 0$, $\alpha=1$: unique — (1). (S) $\Delta
e 0$, $\alpha=1, \gamma=28$: unique solution $x=11, y=-2, z=0$ — (4).

JEE Advanced 2023 Paper 1 · Q14 · Systems of Equations

Solution