If the system of equations
αx + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = β
has infinitely many solutions, then the ordered pair (α, β) is equal to :

Solution

Maths sir · 4 steps

Infinitely many solutions needs D = 0 first. Then the third equation must be a combination of the other two.

1. Condition on D. If D ≠ 0, the system has exactly one solution. So infinitely many solutions forces D = 0, and then every Dₓ, D_y, D_z must also be 0.

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

2. Find alpha. D = 0 gives α = 1. With α = 1, the equations are x + y + z = 5 and x + 2y + 3z = 4, and the third left side is x + 3y + 5z.
3. Find beta. Notice 2(x + 2y + 3z) - (x + y + z) = x + 3y + 5z. For the third equation not to contradict the first two, its right side must be 2(4) - 5 = 3. As a check, Dₓ = |5, 1, 1; 4, 2, 3; β, 3, 5| = β - 3, which is 0 only when β = 3.
4. So. (α, β) = (1, 3), which is option (c).

The idea to keep. For infinitely many solutions, use D = 0 to fix the coefficient. Then make the right sides match the same combination that links the left sides, or set Dₓ = 0.

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