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
- 4x2 – 24x + 27 = 0
- 4x2 – 24x – 27 = 0
- 4x2 – 24x + 27 = 0
- 4x2 + 24x – 27 = 0
The Correct Option is A
Solution and Explanation
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)
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