For real numbers α and β, consider the following system of linear equations :
x + y - z = 2, x + 2y + αz = 1, 2x - y + z = β. If the system has infinite solutions, then α + β is equal to ______________.

Solution

Maths sir · 5 steps

Make the coefficient determinant zero to get α, then force the equations to agree to get β.

1. Condition on alpha. Infinitely many solutions needs the coefficient determinant Δ to be zero. Otherwise the system would have exactly one solution.

   Δ = |1, 1, -1; 1, 2, α; 2, -1, 1| = (2+α) - (1-2α) - (-1-4) = 6 + 3α

2. Solve for alpha. 6 + 3α = 0 gives α = -2. Now the second equation is x + 2y - 2z = 1.
3. Eliminate x. Second minus first gives y - z = -1. Third minus twice the first gives -3y + 3z = β - 4, that is, y - z = (4-β)/3.
4. Make them agree. Both lines describe y - z, so they must match: (4-β)/3 = -1, which gives β = 7. Otherwise the system has no solution. You can check that Δ₂ = Δ₃ = β - 7 also vanishes here.
5. So. α + β = -2 + 7 = 5.

The idea to keep. For infinite solutions, Δ = 0 fixes the coefficient parameter. Then the constant parameter is whatever makes the dependent equations consistent instead of contradictory.

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