Question

If $p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t=\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda -1& \lambda +3\\ {\lambda }^2+1& 2+5\lambda & \lambda -3\\ {\lambda }^2-3& \lambda +4& 3\lambda \end{array}\right|$
then $p$ is equal to

• A. 5
• B. 8
• C. 3
• D. 2

Answer 1

The correct option is B 8

We divide LHS by ${\lambda }^4$ and $C_1$ by ${\lambda }^2$ ,$C_2$ by $\lambda$ and $C_3$ by $\lambda$ on the RHS to obtain ,
$p+q\left(\frac{1}{\lambda }\right)+r{\left(\frac{1}{\lambda }\right)}^2+s{\left(\frac{1}{\lambda }\right)}^3+t{\left(\frac{1}{\lambda }\right)}^4$
$=\left|\begin{array}{ccc}1+3/\lambda & 1-1/\lambda & 1+3/\lambda \\ 1+1/{\lambda }^2& 2/\lambda +5& 1-3/\lambda \\ 1-3/{\lambda }^2& 1+4/{\lambda }^2& 3\end{array}\right|$
taking limit as $\lambda \to \infty$ we get
$p=\left|\begin{array}{ccc}1& 1& 1\\ 1& 5& 1\\ 1& 1& 3\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 0& 4& 0\\ 0& 0& 2\end{array}\right|=8$ ---------$[using{R}_2\to R_2-R_1,R_3\to R_3-R_1]$