Question

Consider the given system of linear equations for variables xx and yy, where kk is a real-valued constant. Which of the following option(s) is/are CORRECT?

x+ky=1x + ky = 1
kx+y=−1kx + y = -1
Choose every correct option
Official answer and solution
Official answer
A and D
From
Official GATE 2025 answer key

Solution

Compute the determinant of the coefficient matrix and check the two singular cases k=1k = 1 and k=−1k = -1.

The determinant of the coefficient matrix [1kk1]\begin{bmatrix} 1 & k \\ k & 1 \end{bmatrix} is Δ=1−k2\Delta = 1 - k^2.

  • If k≠±1k \ne \pm 1: Δ≠0\Delta \ne 0, so the system has a unique solution for infinitely many real values of kk ((C) is false).
  • If k=1k = 1: the equations become x+y=1x + y = 1 and x+y=−1x + y = -1, which are parallel and inconsistent (no solution).
  • If k=−1k = -1: the equations become x−y=1x - y = 1 and −x+y=−1-x + y = -1, which are identical (infinitely many solutions).

Thus, there is exactly one value of kk (k=1k = 1) with no solution (A), and exactly one value of kk (k=−1k = -1) with an infinite number of solutions (D).

Final answer

Options (A), (D)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question
23 of 65
Section
Computer Science and Information Technology

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