Let the system of linear equations
x + 2y + z = 2,
α x + 3y - z = α,
- α x + y + 2z = - α
be inconsistent. Then α is equal to :

Solution

Maths sir · 4 steps

Inconsistent means the determinant is zero and the system still has no solution, so check both.

1. When it fails. A system of three equations has no solution only if Δ = 0 and at least one of Δₓ, Δ_y, Δ_z is not zero. So find where Δ = 0 first.

   Δ = |1, 2, 1; α, 3, -1; -α, 1, 2| = 1(6+1) - 2(2α - α) + 1(α + 3α) = 7 + 2α

2. Set it to zero. 7 + 2α = 0 gives α = -7/2. For every other value of α, Δ ≠ 0 and the system has exactly one solution.
3. Check one more. Replace the x-column with the right-hand side (2, α, -α) to get Δₓ = 2(7) - 2(α) + 1(4α) = 14 + 2α. At α = -7/2 this is 14 - 7 = 7 ≠ 0, so the equations cannot all hold together.
4. So. Δ = 0 and Δₓ ≠ 0, so the system is inconsistent at α = -7/2, which is option (d).

The idea to keep. For no solution, make Δ = 0 first, then check that one of Δₓ, Δ_y, Δ_z is non-zero. If all of them are zero too, the system may have infinitely many solutions instead.

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