Question

Consider the set of all determinants of order $3$ with entries $0$ or $1$ only, Let $B$ be the subset of $A$ consisting of all determinants with value $1$. Let $C$ be the subset of the set of all determinants with value $-1$. Then

• A. $C$ is empty
• B. $B$ has as many elements as $C$
• C. $A=B\cup C$
• D. $B$ has twice as many elements as $C$.

Answer 1

The correct option is B $B$ has as many elements as $C$

We know that the interchange of two adjacent rows (or columns) changes the value of a determine only in sign and not in magnitude. Hence corresponding to every element $△$ of $B$ there is an element $△$ in $C$ obtained by interchanging two adjacent rows (or columns) in $△$. It follows that

$n\left(B\right)\le n\left(C\right)$.

That is, the number of elements in $B$ is less than or equal to the number of elements in $C$.

Similarly $n\left(C\right)\le n\left(B\right)$.

Hence $n\left(B\right)=n\left(C\right)$, that is, $B$ has as many elements as $C$.