Question

$\left| \begin{matrix} x + 1 & x + 2 & x + \lambda \\ x + 2 & x + 3 & x + \mu \\ x + 3 & x + 4 & x + \nu \end{matrix} \right|$ =0, $\lambda,\mu,\nu$ are in A. P. is

• A. An equation whose all roots are real
• B. An identity in x
• C. An equation with only one real root
• D. None of these

Answer 1

Answer: (B)

An identity in x

Answer 2

Let ∆= $\left| \begin{matrix} x + 1 & x + 2 & x + \lambda \\ x + 2 & x + 3 & x + \mu \\ x + 3 & x + 4 & x + \nu \end{matrix} \right|$

Applying $R_{2} \rightarrow R_{2} - \frac{1}{2}\left( R_{1} + R_{3} \right)$

= $\left| \begin{matrix} x + 1 & x + 2 & x + \lambda \\ 0 & 0 & \mu - \frac{\lambda + \nu}{2} \\ x + 3 & x + 4 & x + \nu \end{matrix} \right|$

}{= \left( \mu - \frac{\lambda + \nu}{2} \right)( - 2) = 0}$$ ⇒ 0 = 0 $(\because\lambda,\mu,\nu areinA.P.)$ Hence ∆ is an identify in x.