The number of values of α for which the system of equations :
x + y + z = α
αx + 2αy + 3z = -1
x + 3αy + 5z = 4
is inconsistent, is

Solution

Maths sir · 4 steps

Find where the determinant vanishes, then test that value to see if the equations contradict each other.

1. When trouble starts. If the coefficient determinant Δ ≠ 0, the system has exactly one solution. So it can only be inconsistent at values of α where Δ = 0.

   Δ = |1, 1, 1; α, 2α, 3; 1, 3α, 5| = (10α - 9α) - (5α - 3) + (3α² - 2α) = 3(α - 1)²

2. Only one candidate. 3(α - 1)² = 0 gives only α = 1. It is a repeated root, but it is still a single value of α.
3. Test that value. At α = 1 the equations become x + y + z = 1, x + 2y + 3z = -1 and x + 3y + 5z = 4. Subtracting the first from the second gives y + 2z = -2, and subtracting the second from the third gives y + 2z = 5. Both cannot hold, so the system has no solution.
4. So. The system is inconsistent only for α = 1, which is exactly one value. The answer is (b).

The idea to keep. Δ = 0 only tells you the system is not uniquely solvable. Always put that value back in and check whether the equations contradict each other or agree.

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