Consider the given system of linear equations for variables x and y, where k is a real-valued consta

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Question

Consider the given system of linear equations for variables x and y, where k is a real-valued constant.x + ky = 1kx + y = −1Which of the following option(s) is/are CORRECT?

Accepted Answer

Correct Answer: (A) and (D)Concept — System of Linear Equations: For a 2×2 system Ax = b, the coefficient matrix is:A = [[1, k], [k, 1]], b = [1, −1]Step 1 — Compute the determinant:det(A) = (1)(1) − (k)(k) = 1 − k²Case 1: det(A) ≠ 0 → Unique solutionThis holds when k ≠ 1 and k ≠ −1. For all such values (infinitely many), there is a unique solution. Option C is FALSE (not exactly one value of k gives unique solution — all k except ±1 do).Case 2: det(A) = 0 → k = 1 or k = −1When k = 1: The equations become x + y = 1 and x + y = −1. These are contradictory (same LHS, different RHS). No solution.When k = −1: The equations become x − y = 1 and −x + y = −1. The second equation is just −1 times the first → they are identical. Infinitely many solutions (e.g., x = 1+t, y = t for any t ∈ ℝ).Evaluating each option:Option A (TRUE): Exactly one value of k (k = 1) gives no solution.Option B (FALSE): Only one value (k = 1) gives no solution, not infinitely many.Option C (FALSE): Infinitely many values of k (all k ≠ ±1) give a unique solution.Option D (TRUE): Exactly one value (k = −1) gives infinitely many solutions.Reference: Linear Algebra — Gilbert Strang, Chapter 1 (Systems of Linear Equations).

Consider the given system of linear equations for variables x and y, where k is a real-valued constant.x + ky = 1kx + y = −1Which of the following option(s) is/are CORRECT?

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Consider the given system of linear equations for variables x and y, where k is a real-valued constant.x + ky = 1kx + y = −1Which of the following option(s) is/are CORRECT?

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