Question: If\(p\lambda^{4} + q\lambda^{3} + r\lambda^{2} + s\lambda + t = \left| \begin{matrix} \lambda^{2} + ...

Question

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

value of t is

Answer 1

Solution

Since it is an identity in $\lambda$ so satisfied by every value of $\lambda$ . Now put $\lambda = 0$ in the given equation, we have

0 & - 1 & 3 \\ 1 & 2 & - 4 \\ - 3 & 4 & 0 \end{matrix} \right| = - 12 + 30 = 18$$