Question

If $p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t=\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda -1& \lambda +3\\ \lambda +1& 2-\lambda & \lambda -4\\ \lambda -3& \lambda +4& 3\lambda \end{array}\right|,$ the value of t is

• A. 16
• B. 18
• C. 17
• D. 19

Answer 1

The correct option is B 18

Since it is an identity in $\lambda$ so satisfied by every value of $\lambda$. Now put $\lambda =0$ in the given equation, we have
$t=\left|\begin{array}{ccc}0& -1& 3\\ 1& 2& -4\\ -3& 4& 0\end{array}\right|=-12+30=18.$