Question

Consider the set A 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 A 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 determinant only in sign and not in magnitude. Hence corresponding to every element $\Delta$ of $B$ there is an element ${\Delta }^{\prime }$in $C$ obtained by interchanging two adjacent rows (or columns )in $\Delta .$
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 $C$.
Option D is not correct.
The set A will also have determinants with values 0. Hence, option C is incorrect.
Option A is incorrect as C is not empty.
Hence, option B is correct.