JEE Main 2020 · 5 September · evening · Single correct · +4 / −1
If the system of linear equations
x + y + 3z = 0
x + 3y + k²z = 0
3x + y + 3z = 0
has a non-zero solution (x, y, z) for some k ∈ R, then x + ( y/z ) is equal to :
- (a) 9
- (b) 3
- (c) -9
- (d) -3
Answer and solution
Answer (d) -3
Solution
Maths sir · 4 steps
A non-zero solution means the determinant is zero, but here the first and third equations already give x.
- Condition on k. A homogeneous system has a non-zero solution only when its coefficient determinant is zero.
|1, 1, 3; 1, 3, k²; 3, 1, 3| = (9 - k²) - (3 - 3k²) + 3(1 - 9) = 2k² - 18 = 0
- Find x. This gives k² = 9. Subtract the first equation from the third: (3x + y + 3z) - (x + y + 3z) = 2x = 0, so x = 0.
- Find y over z. With x = 0, the first equation becomes y + 3z = 0, so y/z = -3. Here z ≠ 0, because z = 0 would force y = 0 and give only the zero solution. With k² = 9, the second equation is 3y + 9z = 3(y + 3z) = 0, so it agrees.
- So. x + y/z = 0 + (-3) = -3, which is option (d).
The idea to keep. For a homogeneous system, a non-zero solution needs determinant zero. Then subtract two equations to eliminate variables, and find ratios like y/z instead of exact values.
Answer from the NTA final answer key. Solution by PrepWiser.