JEE Mathematics 2023 Question with Solution

Given: The coefficient matrix of the system is

Find: The sets $S_1$ and $S_2$ for values of $a \in \mathbb{R} - \{0\}$ for which the system has a unique solution and infinitely many solutions.

For a system of three linear equations, a unique solution exists when the determinant of the coefficient matrix is non-zero. Infinitely many solutions can occur only when this determinant is zero and the system is consistent.

From the solution,

Now check when $\Delta = 0$:

So either $a = 0$ or

The quadratic factor cannot be zero because its discriminant is negative:

Hence,

Therefore,

Since the question already excludes $a = 0$, the determinant is non-zero for every allowed value of $a$.

So the system has a unique solution for every $a \in \mathbb{R} - \{0\}$, and it never has infinitely many solutions. Thus,

Therefore, the correct option is D.

Discrepancy note: The provided the solution states "Correct Option is C", but its own working concludes $S_1 = \mathbb{R} - \{0\}$ and $S_2 = \emptyset$, which matches option D.

A system can have infinitely many solutions only if the coefficient determinant becomes zero. Here the determinant reduces to

The allowed values satisfy $a \neq 0$.

So the only remaining possibility for $\Delta = 0$ would be

But this quadratic has no real root because

Hence no real non-zero value of $a$ makes $\Delta = 0$. Therefore the system is never singular on the given domain, so $S_2 = \emptyset$ and consequently $S_1 = \mathbb{R} - \{0\}$. Thus the correct option is D.