Let $\begin{vmatrix} 1+x & x & x^2 \\x & 1+x & x^2 \\x^2 & x & 1+x\end{vmatrix}=ax^5+bx^4 +cx^3 +dx^2 + \lambda x + \mu $

be an identify in x, where $a, b, c, d, \lambda , \mu $ are independent of x. Then, the value of $\lambda $ is,

3

The correct answer is option (4) : 3

Let $Δ =\begin{vmatrix} 1+x & x & x^2 \\x & 1+x & x^2 \\x^2 & x & 1+x\end{vmatrix}$

Applying $C_1→C_1+C_2+C_3$, we get

$Δ=(1+x)^2 \begin{vmatrix} 1 & x & x^2 \\x & 1+x & x^2 \\1 & x & 1+x\end{vmatrix}$

Clearly, $\lambda = \left(\frac{dΔ}{dx}\right)_{x=0}$

We have,

$\frac{dΔ}{dx}=2(1+x) \begin{vmatrix} 1 & x & x^2 \\1 & 1+x & x^2 \\1 & x & 1+x\end{vmatrix}+(1+x)^2\begin{vmatrix} 1 & 1 & x^2 \\1 & 1 & x^2 \\1 & 1 & 1+x\end{vmatrix}+(1+x)^2 \begin{vmatrix} 1 & x & 2x \\1 & 1+x & 2x\\1 & x & 1\end{vmatrix}$

$∴\left(\frac{dΔ}{dx}\right)_{x=0}=2 \begin{vmatrix} 1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{vmatrix}+0+\begin{vmatrix} 1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{vmatrix}=2+1=3.$

Hence, $\lambda = 3$