Q.28 A value of 𝑘 for which the linear equations (𝑘 − 1)𝑥 + 3𝑦 = 0 and 2𝑥 + 𝑘𝑦 = 0
have a non-zero solution is ________.
(A) 1
(B) 2
(C) 3
(D) 4
Finding the Value of k for Non-Trivial Solutions in Linear Equations
The correct answer is (C) 3. For the homogeneous linear equations
(k − 1)x + 3y = 0
2x + ky = 0
to have non-zero (non-trivial) solutions, the determinant of the coefficient matrix must be zero.
Solution Method
The system has non-trivial solutions when the two equations are proportional:
(k − 1) / 2 = 3 / k
Cross-multiplying:
k(k − 1) = 6
Expanding:
k2 − k − 6 = 0
Solving the quadratic equation:
k = (1 ± √(1 + 24)) / 2 = (1 ± 5) / 2
Hence,
- k = 3
- k = −2
Among the given options (1, 2, 3, 4), only k = 3 satisfies the condition.
Determinant Method (Alternative)
The determinant of the coefficient matrix is:
| k − 1 3 |
| 2 k |
Determinant = k(k − 1) − 6
Setting determinant equal to zero:
k(k − 1) − 6 = 0
This leads to the same quadratic equation and solutions as above.
Option Analysis
| Option | k Value | Determinant Check | Has Non-Zero Solution? |
|---|---|---|---|
| (A) | 1 | 1(0) − 6 = −6 ≠ 0 | No |
| (B) | 2 | 2(1) − 6 = 0 | Yes |
| (C) | 3 | 3(2) − 6 = 0 | Yes |
| (D) | 4 | 4(3) − 6 = 6 ≠ 0 | No |
Important Note
Although k = 2 also satisfies the determinant condition mathematically, standard MCQ format requires selecting a single correct option. Among the listed choices, 3 is the correct answer.
For k = 1, the equations reduce to:
0·x + 3y = 0 ⇒ y = 0
2x + y = 0
This system has only the trivial solution (x = 0, y = 0), so it is not acceptable.
