Question

Let $p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t=\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda -1& \lambda +3\\ \lambda +1& -2\lambda & \lambda -4\\ \lambda -3& \lambda +4& 3\lambda \end{array}\right|$ be an identity in $\lambda$, where $p,q,r,s,t$ are constants. Then the value of $t$ is ...............

Answer 1

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

Let $\lambda =0$

$\Rightarrow p\left(0\right)+q\left(0\right)+r\left(0\right)+s\left(0\right)+t=\left|\begin{array}{ccc}0+0& -1& 3\\ 1& 0& -4\\ -3& 4& 0\end{array}\right|$

$\Rightarrow t=\left|\begin{array}{ccc}0& -1& 3\\ 1& 0& -4\\ -3& 4& 0\end{array}\right|$

$\Rightarrow t=0-(-1)(-12)+3\left(4\right)$

$\Rightarrow t=0-12+12$

$\Rightarrow t=0$

Correct answer$=0$