|x+1, x, x; x, x+λ, x; x, x, x+λ²|=9/8(103 x+81), then λ, λ/3 are the roots of the equation :

Solution

Maths sir · 4 steps

Do not expand the whole thing: put x = 0 and the determinant hands you λ straight away.

1. Set x to zero. The equation holds for every x, so it holds at x = 0. The left side becomes a diagonal determinant with entries 1, λ, λ², and its value is their product λ³. The right side becomes 9/8 × 81 = 729/8.

   λ³ = 729/8 = (9/2)³ ⇒ λ = 9/2

2. Check the x term. Expanded fully, the left side is λ³ + x(λ³ + λ² + λ). With λ = 9/2 the coefficient of x is (729 + 162 + 36)/8 = 927/8 = (9 × 103)/8, which matches the right side.
3. Build the quadratic. The roots are λ = 9/2 and λ/3 = 3/2. Their sum is 6 and their product is 27/4, so the equation is x² - 6x + 27/4 = 0.
4. So. Multiplying by 4 gives 4x² - 24x + 27 = 0, which is option (b).

The idea to keep. When a determinant identity holds for all x, substitute a value that kills most entries. Then build the quadratic as x² - (sum)x + product = 0 and watch the signs.

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