Let the system of linear equations
4x + λy + 2z = 0
2x - y + z = 0
μx + 2y + 3z = 0, λ, μ∈R.
has a non-trivial solution. Then which of the following is true?

Solution

Maths sir · 4 steps

A homogeneous system has a non-trivial solution only when its determinant is zero, so factorise it.

1. The condition. All three equations have 0 on the right, so x = y = z = 0 always works. A second, non-trivial solution exists only when the coefficient determinant is zero.

   |4, λ, 2; 2, -1, 1; μ, 2, 3| = 4(-3-2) - λ(6-μ) + 2(4+μ)

2. Expand and factorise. This gives -20 - 6λ + λμ + 8 + 2μ = λμ - 6λ + 2μ - 12. Group the terms as λ(μ - 6) + 2(μ - 6) = (λ + 2)(μ - 6).
3. Set it to zero. (λ + 2)(μ - 6) = 0 holds when μ = 6, whatever λ is, or when λ = -2, whatever μ is.
4. So. Of the options, only μ = 6 with λ ∈ ℝ matches. The answer is (a).

The idea to keep. Homogeneous system with a non-trivial solution means Δ = 0. When Δ has two parameters, try to factorise it: each factor gives a value that works for every value of the other parameter.

Answer from the NTA final answer key. Solution by PrepWiser.