Question

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

• A. $4$
• B. $0$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$0$

$\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda -1& \lambda +3\\ \lambda +1& -2\lambda & \lambda -4\\ \lambda -3& \lambda +4& 3\lambda \end{array}\right|=p{\lambda }^4+q{\lambda }^3+r{\lambda }^2+s\lambda +t$
Substitute $\lambda =0$
$\left|\begin{array}{ccc}0& -1& 3\\ 1& 0& -4\\ -3& 4& 0\end{array}\right|=t$
which is the determinant of skew-symmetric matrix of odd order
Hence, $t=0$