Question

Consider the set A of all determinants of order 3 with entries 0 and 1 only. Let B be the subset of A consisting of all determinants with value 1.
Let C be the subset of the set 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

Answer: (B)

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 $n(B)\le n(C)$.
That is, the number of elements in B is less than or equal to the number of elements in C.
Similarly n (C) $\le$ n (B).
Hence n (B) = n (C), that is, B has many elements as C.