Let S denote the set of all real values of λ such that the system of equations
λ x+y+z=1
x+λ y+z=1
x+y+λ z=1
is inconsistent, then Σλ ∈ S(|λ|²+|λ|) is equal to

Solution

Maths sir · 4 steps

Make the determinant zero, then add all three equations to see which root breaks the system.

1. Need zero determinant. A system with a unique solution is never inconsistent. So Δ, the determinant of the coefficients, must be 0 first.

   Δ = |λ, 1, 1; 1, λ, 1; 1, 1, λ| = λ³ - 3λ + 2 = (λ - 1)²(λ + 2)

2. Check λ = 1. All three equations become x + y + z = 1. That is one plane, so there are infinitely many solutions. The system is consistent, so λ = 1 is not in S.
3. Check λ = −2. Add the three equations: (λ + 2)(x + y + z) = 3. At λ = -2 this reads 0 = 3, which is impossible. So S = {-2}.
4. So. |λ|² + |λ| = 4 + 2 = 6, which is option (d).

The idea to keep. Δ = 0 only gives you the candidates. Test each root in the equations themselves: if they collapse into one equation, the system is consistent, and if they contradict, it is inconsistent.

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