GATE CS 2025 Set 1 Q13 — Linear Algebra: Matrices, Determinants, System of Linear Equations

We are given the system of linear equations in variables $x$ and $y$:

$$\begin{aligned} x + ky &= 1 \\ kx + y &= -1 \end{aligned}$$

We can represent this system in matrix form $Ax = b$ as:

$$\begin{pmatrix} 1 & k \\ k & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

The determinant of the coefficient matrix $A$ is:

$$\det(A) = 1 - k^2$$

Case 1: $\det(A) \neq 0 \implies k \neq \pm 1$

When $k \neq 1$ and $k \neq -1$, the system has a unique (exactly one) solution. Since there are infinitely many real numbers other than $1$ and $-1$, there are infinitely many values of $k$ for which the system has exactly one solution. This makes Option C incorrect.

Case 2: $\det(A) = 0 \implies k = 1$ or $k = -1$

• Subcase 2a: $k = 1$
  Substituting $k = 1$ into the equations: $$\begin{aligned} x + y &= 1 \\ x + y &= -1 \end{aligned}$$ These two equations represent parallel lines that never intersect. Thus, there is no solution. This happens for exactly one value of $k$ (which is $k = 1$). This makes Option A correct and Option B incorrect.
• Subcase 2b: $k = -1$
  Substituting $k = -1$ into the equations: $$\begin{aligned} x - y &= 1 \\ -x + y &= -1 \implies x - y = 1 \end{aligned}$$ Both equations are identical, representing the same line. Thus, there are infinitely many solutions. This happens for exactly one value of $k$ (which is $k = -1$). This makes Option D correct.

Conclusion

• A is CORRECT: Exactly one value of $k$ ($k = 1$) yields no solution.
• D is CORRECT: Exactly one value of $k$ ($k = -1$) yields infinitely many solutions.