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# 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

For every 'det with value 1'($\in B)$ we can find a,get with value -1 by changing the sign of one entry of 1.
Hence there are equal number of elements in B and C.
Hence (b) is the correct option.