Matrices and Determinants 2 Question 28
«««< HEAD
31. Let $p \lambda^{4}+q \lambda^{3}+r \lambda^{2}+s \lambda+t=\begin{array}{ccc}\lambda^{2}+3 \lambda & \lambda-1 & \lambda+3 \ \lambda+1 & -2 \lambda & \lambda-4 \ \lambda-3 & \lambda+4 & 3 \lambda\end{array}$ be an identity in $\lambda$, where $p, q, r, s$ and $t$ are constants. Then, the value of $t$ is…. .
======= ####31. Let $p \lambda^{4}+q \lambda^{3}+r \lambda^{2}+s \lambda+t$
$= \begin{vmatrix}\lambda^{2}+3 \lambda & \lambda-1 & \lambda+3 \\ \lambda+1 & -2 \lambda & \lambda-4 \\ \lambda-3 & \lambda+4 & 3 \lambda\end{vmatrix}$ be an identity in $\lambda$, where $p, q, r, s$ and $t$ are constants. Then, the value of $t$ is…. .
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed
(1981, 2M)
Show Answer
Answer:
Correct Answer: 31. (0)
Solution:
- Given, $\begin{vmatrix}\lambda^{2}+3 \lambda & \lambda-1 & \lambda+3 \\ \lambda+1 & -2 \lambda & \lambda-4 \\ \lambda-3 & \lambda+4 & 3 \lambda\end{vmatrix}$
$ =p \lambda^{4}+q \lambda^{3}+r \lambda^{2}+s \lambda+t $
Thus, the value of $t$ is obtained by putting $\lambda=0$.
$ \begin{aligned} & \Rightarrow \quad\begin{vmatrix} 0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{vmatrix}=t \\ & \Rightarrow \quad t=0 \end{aligned} $
$[\because$ determinants of odd order skew-symmetric matrix is zero]
