The ordered pair (a, b), for which the system of linear equations
3x - 2y + z = b
5x - 8y + 9z = 3
2x + y + az = -1
has no solution, is :

Solution

Maths sir · 4 steps

First make the determinant zero to fix a, then pick b so the equations contradict each other.

1. Need D = 0. If D ≠ 0, the system has exactly one solution. So for no solution, the coefficient determinant must be zero first.

   D = |3, -2, 1; 5, -8, 9; 2, 1, a| = 3(-8a-9) + 2(5a-18) + 1(5+16) = -14a - 42

2. Find a. -14a - 42 = 0 gives a = -3. This alone rules out options (a) and (d).
3. Check D1. Replace the x-column by the right-hand sides: D₁ = |b, -2, 1; 3, -8, 9; -1, 1, -3| = 15b + 0 - 5 = 15b - 5. With D = 0, the system has no solution when D₁ ≠ 0, that is when b ≠ 1/3.
4. So. We need a = -3 and b ≠ 1/3. Among the options only (-3, -1/3) fits, so the answer is (c).

The idea to keep. For no solution, two things must happen: D = 0, and at least one of D₁, D₂, D₃ is non-zero. D = 0 with all of them zero usually means infinitely many solutions.

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