Question

Let $p \lambda^4+q \lambda^3+r \lambda^2+s \lambda+t$ $=\begin{vmatrix}\lambda^2+3 \lambda & \lambda-1 & \lambda-3 \\ \lambda-1 & -2 \lambda & \lambda-4 \\ \lambda-3 & \lambda+4 & 3 \lambda \end{vmatrix}$
where $p, q, r, s$ and $t$ are constants. Then value of $t$ is

• A. 0
• B. -1
• C. 2
• D. 3

Answer 1

Answer: (A)

Putting $\lambda=0$, we obtain
$t=\begin{vmatrix}0 & -1 & 3 \\1 & 0 & -4 \\-3 & 4 & 0\end{vmatrix}=0$
as it is a skew symmetric determinant of odd order.