Question

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

• A. $16$
• B. $25$
• C. $18$
• D. $20$

Answer 1

The correct option is C $18$

We have,
$\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda -1& \lambda +3\\ \lambda +1& 2-\lambda & \lambda -4\\ \lambda -3& \lambda +4& 3\lambda \end{array}\right|=p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t$
Putting $\lambda =0$ on both sides, we get
$\Rightarrow \left|\begin{array}{ccc}0& -1& 3\\ 1& 2& -4\\ -3& 4& 0\end{array}\right|=t$
$\Rightarrow 0(0-(-16\left)\right)+1(0-12)+3(4-(-6\left)\right)=t$
$\Rightarrow t=-12+30=18$