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Q. If $ f(x)=\left| \begin{matrix} x-3 & 2{{x}^{2}}-18 & 3{{x}^{3}}-81 \\ x-5 & 2{{x}^{2}}-50 & 4{{x}^{3}}-500 \\ 1 & 2 & 3 \\ \end{matrix} \right|, $ then $ f(1).f(3)+f(3).f(5)+f(5).f(1) $ is equal to:

KEAMKEAM 2005

Solution:

$ \because $ $ f(1)=\left| \begin{matrix} -2 & -16 & -78 \\ -4 & -48 & 496 \\ 1 & 2 & 3 \\ \end{matrix} \right| $
$ f(3)=\left| \begin{matrix} 0 & 0 & 0 \\ -2 & -32 & -392 \\ 1 & 2 & 3 \\ \end{matrix} \right|=0 $
and $ f(5)=\left| \begin{matrix} 2 & 32 & 294 \\ 0 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|=0 $
$ \therefore $ $ f(1).f(3)+f(3)-f(5)+f(5).f(1) $
$ =f(1).0+0+f(1).0 $ $ =0=f(3) $