Let $p \lambda^{4}+q \lambda^{3}+r \lambda^{2}+s \lambda+t=\left|\begin{array}{ccc}\lambda^{2}+3 \la

Question

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

Professor PC Thomas Classes · Answer

Expanding, $p \lambda^{4}+q \lambda^{3}+r \lambda^{2}+s \lambda+t$ $=\left(\lambda^{2}+3 \lambdaight)\left[-6 \lambda^{2}-\left(\lambda^{2}-16ight)ight]-(\lambda-1)[(\lambda+$ 1) $(3 \lambda)-(\lambda-3)(\lambda-4)]+(\lambda+3)[(\lambda+1)(\lambda$ $+4)+2 \lambda(\lambda-3)]$\ Independent term $=t=(1)(-12)+3(4)$\ $=-12+12=0 \quad 	herefore \quad t=0$\