JEE Mathematics 2023 Question with Solution

Given:

$$\begin{vmatrix} x + 1 & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda^2 \end{vmatrix} = \frac{9}{8}(103x + 81)$$

Find: The quadratic equation whose roots are $\lambda$ and $\frac{\lambda}{3}$.

From the solution, substitute $x = 0$. Then the determinant becomes

$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{vmatrix} = \frac{9}{8}(81)$$

so the determinant equals

$$1 \cdot \lambda \cdot \lambda^2 = \lambda^3$$

Hence,

$$\lambda^3 = \frac{9}{8} \cdot 81 = \frac{729}{8}$$

The solution states that the determinant leads to the quadratic

$$4x^2 - 24x + 27 = 0$$

and also states that the roots are $\frac{9}{2}$ and $\frac{3}{2}$. These two roots are indeed the roots of that quadratic because

$$4x^2 - 24x + 27 = 4\left(x - \frac{9}{2}\right)\left(x - \frac{3}{2}\right)$$

The solution directly gives the roots as $\frac{9}{2}$ and $\frac{3}{2}$. A quadratic with these roots is

$$(x - \tfrac{9}{2})(x - \tfrac{3}{2}) = 0$$

Multiplying by $4$,

$$4x^2 - 24x + 27 = 0$$

However, the same the solution labels the correct option as A, while this equation matches option C. Since the worked equation is the primary source, the defensible answer is A only by the page label, but the equation itself matches C.