If \(\begin{vmatrix} x + 4 & 2x & 2x\\ 2x & x + 4 & 2x\\ 2x & 2x & x + 4\end{vmatrix} = \lambda(4 - x)^2\), then value of \(\lambda\) is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\lambda = 5x + 4\)
Every row adds up to 5x + 4, so the properties of determinants give the factor quickly.
Applying \(R_1 \to R_1 + R_2 + R_3\) and taking 5x + 4 out of the first row:
\( \Delta = (5x + 4)\begin{vmatrix} 1 & 1 & 1\\ 2x & x + 4 & 2x\\ 2x & 2x & x + 4\end{vmatrix} \)
Applying \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\):
\( \Delta = (5x + 4)\begin{vmatrix} 1 & 0 & 0\\ 2x & 4 - x & 0\\ 2x & 0 & 4 - x\end{vmatrix} \)
Expanding along the first row:
\(\Delta = (5x + 4)(4 - x)^2\)
Comparing with \(\lambda(4 - x)^2\):
\(\lambda = 5x + 4\)
Shortcut: Put simple values of x. At x = 0 the determinant is 4 × 4 × 4 = 64 = 16λ, so λ = 4: only Options (2) and (3) give 4. At x = 1 the determinant is 5(25 − 4) − 2(10 − 4) + 2(4 − 10) = 81 = 9λ, so λ = 9, which fits 5x + 4 and not 4 − x.