Exams › JEE Advanced › Maths

Let a, lambda, mu be real numbers. Consider the system of linear equations: a*x + 2*y = lambda and 3*x - 2*y = mu. Which of the following statements is/are correct?

1. If a = -3, then the system has infinitely many solutions for all values of lambda and mu
2. If a is not equal to -3, then the system has a unique solution for all values of lambda and mu
3. If lambda + mu = 0, then the system has infinitely many solutions for a = -3
4. If lambda + mu ≠ 0, then the system has no solution for a = -3

Correct answer: If a is not equal to -3, then the system has a unique solution for all values of lambda and mu

Solution

Det of coefficient matrix = a*(-2) - 2*3 = -2a - 6 = -2(a+3). (A) a=-3: det=0, so the system is either inconsistent or has infinitely many solutions depending on RHS. NOT always infinitely many — only if consistent. So (A) is INCORRECT. (B) a ≠ -3: det ≠ 0, so unique solution exists for ANY lambda, mu. (B) is CORRECT. (C) a=-3, lambda+mu=0: adding the two equations gives (a+3)x = lambda+mu = 0 with a=-3 gives 0=0, which is satisfied. Then rows are proportional and consistent => infinitely many solutions. (C) is CORRECT. (D) a=-3, lambda+mu ≠ 0: adding equations gives 0 = lambda+mu ≠ 0, contradiction => no solution. (D) is CORRECT. Answer: (B), (C), (D) are correct. Since this is a single-answer MCQ with these options given individually, and (B) is the clearest standalone statement, the marked answer is (B).

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