Let λ, μ ∈ R. If the system of equations
3 x+5 y+λ z=3 ; 7 x+11 y-9 z=2 ; 97 x+155 y-189 z=μ
has infinitely many solutions, then μ+2 λ is equal to :

Solution

Maths sir · 4 steps

For infinitely many solutions, make the third equation a mix of the first two and match every coefficient.

1. Why a combination. The first two equations are independent, because 3/7 ≠ 5/11. For infinitely many solutions, the third equation must add nothing new, so it must equal a times the first plus b times the second.
2. Match x and y. From the x and y coefficients, 3a + 7b = 97 and 5a + 11b = 155. Multiply the first by 5 and the second by 3, then subtract: 35b - 33b = 485 - 465, so b = 10. Then 3a = 97 - 70, so a = 9.

   R₃ = 9R₁ + 10R₂

3. Match z and constants. For the z coefficient, 9λ + 10(-9) = -189, so 9λ = -99 and λ = -11. For the right-hand side, μ = 9(3) + 10(2) = 47.
4. So. μ + 2λ = 47 - 22 = 25, which is option (b).

The idea to keep. Infinitely many solutions in three variables means one equation is a combination of the others. Fix the multipliers from the cleanest coefficients first, then force the rest to agree.

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