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 $A$ consisting 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 $C$

Explanation for the correct options:

Given,

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 $1$.

$C$ be the subset of $A$ consisting of all determinants with value $-1$.

By interchanging any rows or columns then sign of the determinant will change.

Therefore, $n\left(B\right)=n\left(C\right)$

The number of elements of $B$ is equal to the number of elements in $C$

Hence, option (B) is the correct answer