JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
The values of λ and μ for which the system of linear equations
x + y + z = 2
x + 2y + 3z = 5
x + 3y + λz = μ
has infinitely many solutions are, respectively:
- (a) 6 and 8
- (b) 5 and 8
- (c) 5 and 7
- (d) 4 and 9
Answer and solution
Answer (b) 5 and 8
Solution
Maths sir · 4 steps
Subtract the equations in turn: the last two reduced equations must become one and the same.
- Determinant zero. For infinitely many solutions the coefficient determinant must be zero. Expanding along the first row gives 1(2λ-9)-1(λ-3)+1(3-2)=λ-5, so λ=5.
- Eliminate x. Subtract the first equation from the second: y+2z=3. Subtract the second from the third: y+(λ-3)z=μ-5.
y+2z=3 and y+(λ-3)z=μ-5
- Make them match. These two must be the same equation, not just parallel ones. So λ-3=2 and μ-5=3, which gives λ=5 and μ=8.
- So. The system has infinitely many solutions for λ=5, μ=8, which is option (b).
The idea to keep. Determinant zero only rules out a unique solution. To get infinitely many you also need the equations to agree, otherwise they contradict each other and there is no solution at all.
Answer from the NTA final answer key. Solution by PrepWiser.