If the system of equations
2 x+3 y-z=5 ; ; x+α y+3 z=-4 ; ; 3 x-y+β z=7
has infinitely many solutions, then 13 α β is equal to :

Solution

Maths sir · 4 steps

For infinitely many solutions, D and every Dₓ, D_y, D_z must vanish, so pick the two that are easiest.

1. Condition. A 3 × 3 system has infinitely many solutions only when D = 0 and also Dₓ = D_y = D_z = 0. Here D_z contains only α and D_y contains only β, so use those two first.

   D_z = |2, 3, 5; 1, α, -4; 3, -1, 7| = 2(7α - 4) - 3(7 + 12) + 5(-1 - 3α) = -α - 70

2. Find alpha and beta. D_z = 0 gives α = -70. Next, D_y uses the columns (2, 1, 3), (5, -4, 7) and (-1, 3, β), which gives 2(-4β - 21) - 5(β - 9) - (7 + 12) = -13β - 16. Setting D_y = 0 gives β = -16/13.
3. Check D. Expanding gives D = 2αβ + 3α - 3β + 34. With these values, 2240/13 - 210 + 48/13 + 34 = 176 - 176 = 0, so D = 0 as required.
4. So. 13αβ = 13 × (-70) × (-16/13) = 1120, which is option (b).

The idea to keep. Infinitely many solutions means D and all of Dₓ, D_y, D_z are zero. Look for the determinants that hold only one unknown each and solve those first.

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