Question

Let $\left|\begin{array}{ccc}1+x& x& x^2\\ x& 1+x& x^2\\ x^2& x& 1+x\end{array}\right|=a{x}^5+b{x}^4+c{x}^3+d{x}^2+\lambda x+\mu$ be an identity in $x$, where $a,b,c,d,$ $\lambda ,$ $\mu$ are independent of $x$. Then, the value of $\lambda$ is

• A. $3$
• B. $2$
• C. $4$
• D. None of these

Answer 1

The correct option is A $3$

$\left|\begin{array}{ccc}1+x& x& x^2\\ x& 1+x& x^2\\ x^2& x& 1+x\end{array}\right|=a{x}^5+b{x}^4+c{x}^3+d{x}^2+\lambda x+\mu$
Substituting $x=0$
$\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=\mu$
$\Rightarrow \mu =1$
So, $\left|\begin{array}{ccc}1+x& x& x^2\\ x& 1+x& x^2\\ x^2& x& 1+x\end{array}\right|=a{x}^5+b{x}^4+c{x}^3+d{x}^2+\lambda x+1$
Differentiating w.r.t both sides, we get
$\left|\begin{array}{ccc}1& 1& 2x\\ x& 1+x& x^2\\ x^2& x& 1+x\end{array}\right|+\left|\begin{array}{ccc}1+x& x& x^2\\ 1& 1& 2x\\ x^2& x& 1+x\end{array}\right|+\left|\begin{array}{ccc}1+x& x& x^2\\ x& 1+x& x^2\\ 2x& 1& 1\end{array}\right|=5a{x}^4+4b{x}^3+3c{x}^2+2dx+\lambda$
Substituting $x=0$
$\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 1& 1\end{array}\right|=\lambda$
$\Rightarrow \lambda =1+1+1=3$