JEE Main 2021 · 20 July · evening · Single correct · +4 / −1
The value of k ∈R, for which the following system of linear equations
3x - y + 4z = 3,
x + 2y - 3z = -2
6x + 5y + kz = -3,
has infinitely many solutions, is :
- (a) 3
- (b) -5
- (c) 5
- (d) -3
Answer and solution
Answer (b) -5
Solution
Maths sir · 4 steps
Set the coefficient determinant to zero, then check that the third equation is a mix of the first two.
- Condition needed. Infinitely many solutions need the coefficient determinant D to be zero. That alone is not enough, so we also check that the equations do not contradict each other.
D = |3, -1, 4; 1, 2, -3; 6, 5, k| = 3(2k+15) + (k+18) + 4(5-12) = 7k + 35
- Solve for k. 7k + 35 = 0 gives k = -5.
- Check consistency. With k = -5, take 1 times the first equation plus 3 times the second. The left side is (3+3)x + (-1+6)y + (4-9)z = 6x + 5y - 5z, and the right side is 3 + 3(-2) = -3. That is exactly the third equation, so it adds no new condition.
- So. Only two independent equations remain in three unknowns, giving infinitely many solutions. The answer is k = -5, option (b).
The idea to keep. For infinitely many solutions, D = 0 comes first. Then confirm one equation is a combination of the others, right side included, or the system has no solution.
Answer from the NTA final answer key. Solution by PrepWiser.