1. Key Concepts and Formulas

• Determinant Properties: Applying elementary row or column operations (e.g., $R_i \to R_i - R_j$ or $C_i \to C_i - C_j$) can simplify a determinant without changing its value, making expansion easier.

• Determinant of a specific form: For a $3 \times 3$ determinant where all off-diagonal elements are $x$, and diagonal elements are $x+a, x+b, x+c$, the determinant can be generally expressed as $x(ab+bc+ca) + abc + x^3 + (a+b+c)x^2 - x^3 - (a+b+c)x^2 = x(ab+bc+ca) + abc$. A more general form of determinant $\left|\begin{array}{ccc}x+a & x & x \\ x & x+b & x \\ x & x & x+c\end{array}\right|$ is $abc + x(ab+bc+ca + a+b+c) + x^2(a+b+c) + x^3 - x^3 - x^2(a+b+c) = abc + x(ab+bc+ca)$. A simpler expansion is $x(ab+bc+ca) + abc$. Let's re-verify the earlier derived formula for $D = \left|\begin{array}{ccc}x+a & x & x \\ x & x+b & x \\ x & x & x+c\end{array}\right|$ as $x(a+b+c) + abc$. No, this is incorrect. The correct general formula for this specific determinant form is $abc + x(ab+bc+ca)$. My previous detailed expansion was $x(\lambda^3+\lambda^2+\lambda) + \lambda^3$. Let's check the general formula for $D = \left|\begin{array}{ccc}x+a & x & x \\ x & x+b & x \\ x & x & x+c\end{array}\right|$ Using $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $D = \left|\begin{array}{ccc}x+a & x & x \\ -a & b & 0 \\ -a & 0 & c\end{array}\right|$ Expanding along $R_1$: $D = (x+a)(bc - 0) - x(-ac - 0) + x(0 - (-ab))$ $D = (x+a)bc + xac + xab$ $D = xbc + abc + xac + xab$ $D = x(ab+bc+ca) + abc$. This is the correct general form.

• Polynomial Identity: If two polynomials are equal for all values of their variable, then the coefficients of corresponding powers of the variable must be equal. This allows us to equate coefficients to solve for unknown parameters.

• Forming a Quadratic Equation: If $\alpha$ and $\beta$ are the roots of a quadratic equation, the equation can be written as $x^2 - (\alpha+\beta)x + \alpha\beta = 0$.

2. Step-by-Step Solution

Step 1: Evaluate the determinant. We are given the determinant: $D = \left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|$ This determinant is of the form $\left|\begin{array}{ccc}x+a & x & x \\ x & x+b & x \\ x & x & x+c\end{array}\right|$, where $a=1$, $b=\lambda$, and $c=\lambda^2$.

We can simplify the determinant by applying row operations: Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $D = \left|\begin{array}{ccc}x+1 & x & x \\ x-(x+1) & (x+\lambda)-x & x-x \\ x-(x+1) & x-x & (x+\lambda^{2})-x\end{array}\right|$ $D = \left|\begin{array}{ccc}x+1 & x & x \\ -1 & \lambda & 0 \\ -1 & 0 & \lambda^{2}\end{array}\right|$ Now, expand the determinant along the first row: $D = (x+1) \left|\begin{array}{cc}\lambda & 0 \\ 0 & \lambda^{2}\end{array}\right| - x \left|\begin{array}{cc}-1 & 0 \\ -1 & \lambda^{2}\end{array}\right| + x \left|\begin{array}{cc}-1 & \lambda \\ -1 & 0\end{array}\right|$ $D = (x+1)(\lambda \cdot \lambda^2 - 0 \cdot 0) - x((-1) \cdot \lambda^2 - 0 \cdot (-1)) + x((-1) \cdot 0 - \lambda \cdot (-1))$ $D = (x+1)(\lambda^3) - x(-\lambda^2) + x(\lambda)$ $D = x\lambda^3 + \lambda^3 + x\lambda^2 + x\lambda$ $D = x(\lambda^3 + \lambda^2 + \lambda) + \lambda^3$

Step 2: Equate coefficients with the given expression. We are given that $D = \frac{9}{8}(103 x+81)$. $D = \frac{9 \cdot 103}{8} x + \frac{9 \cdot 81}{8}$ $D = \frac{927}{8} x + \frac{729}{8}$ By comparing the coefficients of $x$ and the constant terms from our derived determinant and the given expression:

1. Comparing constant terms: $\lambda^3 = \frac{729}{8}$ $\lambda^3 = \left(\frac{9}{2}\right)^3$ Since $\lambda$ is typically considered a real number in such problems, we take the real root: $\lambda = \frac{9}{2}$

2. Comparing coefficients of $x$: $\lambda^3 + \lambda^2 + \lambda = \frac{927}{8}$ Let's substitute the value of $\lambda = \frac{9}{2}$ to verify this consistency: $\left(\frac{9}{2}\right)^3 + \left(\frac{9}{2}\right)^2 + \frac{9}{2} = \frac{729}{8} + \frac{81}{4} + \frac{9}{2}$ To add these fractions, find a common denominator, which is 8: $= \frac{729}{8} + \frac{81 \cdot 2}{4 \cdot 2} + \frac{9 \cdot 4}{2 \cdot 4}$ $= \frac{729}{8} + \frac{162}{8} + \frac{36}{8}$ $= \frac{729 + 162 + 36}{8} = \frac{927}{8}$ This matches the coefficient of $x$ from the given expression. Thus, our value of $\lambda = \frac{9}{2}$ is correct.

Step 3: Form the quadratic equation. The problem states that $\lambda$ and $\frac{\lambda}{3}$ are the roots of the equation. The first root is $\alpha = \lambda = \frac{9}{2}$. The second root is $\beta = \frac{\lambda}{3} = \frac{9/2}{3} = \frac{9}{6} = \frac{3}{2}$.

Now, we find the sum and product of these roots: Sum of roots: $\alpha + \beta = \frac{9}{2} + \frac{3}{2} = \frac{12}{2} = 6$. Product of roots: $\alpha \beta = \frac{9}{2} \cdot \frac{3}{2} = \frac{27}{4}$.

The quadratic equation is given by $x^2 - (\alpha+\beta)x + \alpha\beta = 0$: $x^2 - 6x + \frac{27}{4} = 0$ To eliminate the fraction, multiply the entire equation by 4: $4(x^2 - 6x + \frac{27}{4}) = 0 \cdot 4$ $4x^2 - 24x + 27 = 0$

This derived equation corresponds to option (B).

3. Common Mistakes & Tips

• Determinant Expansion Errors: Be careful with signs when expanding determinants, especially with negative terms or cofactors. Double-check your calculations.
• Algebraic Simplification: Ensure careful handling of fractions and combining like terms. Mistakes in basic algebra can propagate throughout the solution.
• Equating Coefficients: Remember that for polynomial identity, coefficients of each power of the variable must match. Don't just compare constant terms or highest power terms.

4. Summary

The problem required evaluating a $3 \times 3$ determinant, which simplified to a linear polynomial in $x$. By equating the coefficients of this polynomial with the given right-hand side expression, we determined the value of $\lambda$. Once $\lambda$ was found, we calculated the two roots $\lambda$ and $\lambda/3$ and then used the sum and product of roots to form the desired quadratic equation. The derived equation is $4x^2 - 24x + 27 = 0$.

5. Final Answer

The final answer is \boxed{A}.