A) \[f(1)\]
B) \[f(3)\]
C) \[f(1)+f(3)\]
D) \[f(1)+f(5)\]
E) \[f(1)+f(3)+f(5)\]
Correct Answer: B
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 \[f(5)=\left| \begin{matrix} 2 & 32 & 294 \\ 0 & 0 & 0 \\ 1 & 2 & 3 \\ \end{matrix} \right|=0\] \[\therefore \] \[f(1).f(3)+f(3)-f(5)+f(5).f(1)\] \[=f(1).0+0+f(1).0\] \[=0=f(3)\]
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