Browse Questions

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

$(a)\;t=1\qquad(b)\;t=0\qquad(c)\;t=2\qquad(d)\;t=3$

$p\lambda^4+q\lambda^3+r\lambda^2+s\lambda+t=\begin{vmatrix}\lambda^3+3\lambda&\lambda-1&\lambda+3\\\lambda+1&-2\lambda&\lambda-4\\\lambda-3&\lambda+4&3\lambda\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\lambda^3&\lambda-1&\lambda+3\\\lambda&-2\lambda&\lambda-4\\\lambda&\lambda+4&3\lambda\end{vmatrix}+\begin{vmatrix}3\lambda&\lambda-1&\lambda+3\\1&-2\lambda&\lambda-4\\-3&\lambda+4&3\lambda\end{vmatrix}$
$\Rightarrow \lambda \begin{vmatrix}\lambda^2&\lambda-1&\lambda+3\\1&-2\lambda&\lambda-4\\1&\lambda+4&3\lambda\end{vmatrix}+3\lambda(-6\lambda^2-\lambda^2+4)-(\lambda-1)(3\lambda+3\lambda-12)+(\lambda+3)(\lambda+4-6\lambda)$
$\Rightarrow$(multiple of $\lambda$)-$21\lambda^3-12\lambda-6\lambda^2+12\lambda+6\lambda-12+4\lambda-5\lambda^2+12-15\lambda$
$\Rightarrow$(multiple of $\lambda$)-$21\lambda^3-11\lambda^2+17\lambda$
$\Rightarrow$ Constant term (independent of $\lambda$) is zero on RHS
$\Rightarrow$ equating two sides we get $t=0$
Hence (b) is the correct answer.