The system of linear equations
λx + 2y + 2z = 5
2λx + 3y + 5z = 8
4x + λy + 6z = 10 has

Solution

Maths sir · 4 steps

Find where the determinant is zero first. Only those values of λ can stop the solution being unique.

1. Expand the determinant. Expand along the first row: λ(18-5λ) - 2(12λ-20) + 2(2λ²-12) = -λ² - 6λ + 16.

   D = |λ, 2, 2; 2λ, 3, 5; 4, λ, 6| = -(λ+8)(λ-2)

2. Read off the cases. D = 0 only at λ = 2 and λ = -8. For every other value, including λ = 8 where D = -96, the system has a unique solution.
3. Check the value 2. With λ = 2 the equations become 2x+2y+2z=5, 4x+3y+5z=8 and 4x+2y+6z=10. Second minus twice the first gives z-y=-2. Third minus twice the first gives -2y+2z=0, so z-y=0.
4. So. z-y cannot be both -2 and 0, so the equations contradict each other. At λ = 2 there is no solution, which is option (b).

The idea to keep. First solve D = 0 to find the doubtful values. At each one, check whether the equations agree or contradict each other. D = 0 alone never tells you which.

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