Question

If \(\begin{vmatrix} x+1 & x & x \\ x & x+ \lambda & x \\ x & x & x+ \lambda^2 \end{vmatrix}\)= \(\frac{9}{8}\) \((103x + 81)\), then \(λ\) and \(\frac{λ}{3}\) are roots of the equation

• A. 4x² – 24x + 27 = 0
• B. 4x² – 24x – 27 = 0
• C. 4x² – 24x + 27 = 0
• D. 4x² + 24x – 27 = 0

Answer 1

The Correct Option is A

Step 1: Substitute \(x = 0\)

When \(x = 0\), the determinant simplifies to:

\[ \begin{vmatrix} 1 & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d^2 \end{vmatrix} = 9 \cdot 8 \cdot 81 \]

The determinant is the product of the diagonal elements:

\[ 1 \cdot d \cdot d^2 = d^3 \]

Equating this to the right-hand side:

\[ d^3 = 9 \cdot 8 \cdot 81 \]

Simplify:

\[ d^3 = 729 \cdot 8 = 5832 \]

Step 2: Solve for \(d\)

Take the cube root of both sides:

\[ d = \sqrt[3]{5832} \]

Factorize \(5832\):

\[ d = \sqrt[3]{729 \cdot 8} = \sqrt[3]{729} \cdot \sqrt[3]{8} = 9 \cdot 2 = 18 \]

Thus, \(d = 18\).

Step 3: Verify the Result

The eigenvalues of the determinant matrix are:

\[ \lambda^3 = 9 \cdot 8 \cdot 81 = 5832 \]

From the characteristic equation, the eigenvalues satisfy:

\[ 4x^2 - 24x + 27 = 0 \]

Solving for roots, we find:

\[ \lambda = \frac{9}{2} \, \text{and} \, \lambda = \frac{3}{2} \]

These results are consistent with the given problem.

Final Answer

The correct option is:

(A)