If $p\lambda^{4} + q\lambda^{3} + q\lambda^{2} + s\lambda + t = \left| \begin{matrix} b^{2} + c^{2} & a^{2} + \lambda & a^{2} + \lambda \ b^{2} + \lambda & c^{2} + a^{2} & b^{2} + \lambda \ c^{2} + \lambda & c^{2} + \lambda & a^{2} + b^{2} \end{matrix} ight|$ is an identity in $\lambda,$ where $p,q,r,s,t$ are constants, then the value of $t$ is

Question

If $p\lambda^{4} + q\lambda^{3} + q\lambda^{2} + s\lambda + t = \left| \begin{matrix} b^{2} + c^{2} & a^{2} + \lambda & a^{2} + \lambda \ b^{2} + \lambda & c^{2} + a^{2} & b^{2} + \lambda \ c^{2} + \lambda & c^{2} + \lambda & a^{2} + b^{2} \end{matrix} ight|$ is an identity in $\lambda,$ where $p,q,r,s,t$ are constants, then the value of $t$ is

Dr. Abubacker · Accepted Answer

(d)
On putting $\lambda = 0$, we get
$t = \left| \begin{matrix}
b^{2} + c^{2} & a^{2} & a^{2} \
b^{2} & {c^{2} + a}^{2} & b^{2} \
c^{2} & c^{2} & a^{2} + b^{2}
\end{matrix} ight| = 4a^{2}b^{2}c^{2}$
Clearly, it depends on $a,b,c$.

Explanation

Reviewed by Dr. Abubacker

Want more practice?