If the system of linear equations
x-2 y+z=-4 ; 2 x+α y+3 z=5 ; 3 x-y+β z=3
has infinitely many solutions, then 12 α+13 β is equal to

Solution

Maths sir · 5 steps

Start with D_z and D_y: each has only one unknown, so α and β come out in one line each.

1. Condition. For three equations in three unknowns to have infinitely many solutions, you need D = 0 and also Dₓ = D_y = D_z = 0. D_z and D_y each contain only one unknown, so use those first.

   D_z = |1, -2, -4; 2, α, 5; 3, -1, 3| = (3α + 5) + 2(6 - 15) - 4(-2 - 3α) = 15α - 5

2. Find alpha. 15α - 5 = 0 gives α = 1/3.
3. Find beta. Replace the y-column by the constants -4, 5, 3. Then D_y = (5β - 9) + 4(2β - 9) + (6 - 15) = 13β - 54. Setting D_y = 0 gives β = 54/13.
4. Check D. D = αβ - 3α + 4β - 17 = 18/13 - 1 + 216/13 - 17 = 18 - 18 = 0. The values are consistent.
5. So. 12α + 13β = 12 × 1/3 + 13 × 54/13 = 4 + 54 = 58, which is option (d).

The idea to keep. For infinitely many solutions, every one of D, Dₓ, D_y, D_z must be zero. First pick the determinant with the fewest unknowns. It usually contains just one, which you can solve directly.

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