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Let $A$ be the set of all $3 \times 3$ determinants with entries 0 or 1 only and $B$ be the subset of $A$ consisting of all determinants with value 1 . If $C$ is the subset of $A$ consisting of all determinants with value -1 , then
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The correct answer is:
$n(B)=n(C)$
We know that the interchange of two adjacent rows (columns) changes the value of a determinant only in sign but not in magnitude.
Hence, corresponding to every element $\Delta$ at $B$ there is an element $\Delta^{\prime}$ in $C$ obtained by inter changing two adjacent rows (columns) in $\Delta$. It follows that $n(B) \leq n(c)$, i.e. the number of elements in $B$ is less than an equal to the number of elements in $C$.
Similarly, $\quad n(C) \leq n(B)$
Hence, $\quad n(B)=n(C)$
i.e. $B$ has an many elements as $C$.
Hence, corresponding to every element $\Delta$ at $B$ there is an element $\Delta^{\prime}$ in $C$ obtained by inter changing two adjacent rows (columns) in $\Delta$. It follows that $n(B) \leq n(c)$, i.e. the number of elements in $B$ is less than an equal to the number of elements in $C$.
Similarly, $\quad n(C) \leq n(B)$
Hence, $\quad n(B)=n(C)$
i.e. $B$ has an many elements as $C$.
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