Let A be the set of all determinants of order 3 with entries 0 or 1 only, B is the subset of A consisting of all determinants with value 1, and C is the subset consisting of all determinants with value -1. Then if $n (B)$ and $n (C)$ denote the number of elements in B and C, respectively, we have

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$n (B) = 2 n (C)$

$n (B) = n (C)$

$A = B \cup C$

$C = ϕ$

answer is B.

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Detailed Solution

Since, C cannot be the empty set, Hence
$- 1 = | \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} | \in C$
We also have
$| \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{matrix} | = 2$
So $A \neq B \cup C$ . In general, the determinant
$Δ = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} |$
$Δ = a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} - a_{13} a_{31} a_{22}$
With the $a' s$ being 0 or 1, equals 1 only if $a_{11} a_{22} a_{33} = 1$ and the remaining terms are zero; if $a_{12} a_{23} a_{31} = 1$ and the remaining terms are zero; or if $a_{13} a_{21} a_{32} = 1$ and the remaining terms are zero. Since there are three similar relations for determinants that equal -1, we must have $n (B) = n (C)$