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. $1$

Answer 1

The correct option is A $3$

Given, $\left|\begin{array}{ccc}1+x& x& x^2\\ x& 1+x& x^2\\ x^2& x& 1+x\end{array}\right|=a{x}^5+b{x}^4+c{x}^3+d{x}^2+\lambda x+\mu$
Put $x=0$
$\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=\mu$
$\Rightarrow \mu =1$
Differentiating both sides,
$\left|\begin{array}{ccc}1& 1& 2x\\ x& 1+x& x^2\\ x^2& x& 1+x\end{array}\right|+\left|\begin{array}{ccc}1+x& x& x^2\\ 1& 1& 2x\\ x^2& x& 1+x\end{array}\right|+\left|\begin{array}{ccc}1+x& x& x^2\\ x& 1+x& x^2\\ 2x& 1& 1\end{array}\right|=5a{x}^4+4b{x}^3+3c{x}^2+2dx+\lambda$
Put $x=0$
$\lambda =\left|\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right|+\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 1& 1\end{array}\right|$
$=1+1+1=3$