If the system of equations
x+y+z=2
2x+4y–z=6
3x+2y+λz=μ
has infinitely many solutions, then

Solution

Maths sir · 4 steps

Infinitely many solutions: the third equation must be a combination of the first two, right side included.

1. Determinant is zero. Infinitely many solutions needs Δ = 0. Expanding along the first row gives Δ = (4λ + 2) - (2λ + 3) + (4 - 12) = 2λ - 9, so λ = 9/2.
2. Find the combination. Write the third equation as a times the first plus b times the second. Matching the x and y coefficients gives a + 2b = 3 and a + 4b = 2, so b = -1/2 and a = 4. Check the z coefficient: 4(1) - 1/2(-1) = 9/2 = λ, as required.

   μ = 4(2) - 1/2(6) = 8 - 3 = 5

3. Why match the constant. If μ were anything other than 5, the third equation would contradict the first two and there would be no solution at all. With μ = 5 it adds nothing new, so one free variable remains and the solutions are infinite.
4. So. With λ = 9/2 and μ = 5, 2λ + μ = 9 + 5 = 14. The answer is (c).

The idea to keep. Δ = 0 alone allows either no solution or infinitely many. To get infinitely many, the dependent equation must match the combination on the right-hand side too.

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