Question

Let A be the set of all 3 $\times$ 3 symmetric matrices all of whose entries are either 0 or 1, five of these entries are 1 and four of them are zero. The number of matrices in A is

• A. 12
• B. 9
• C. 6
• D. 3

Answer 1

The correct option is A 12

Each element of set A is 3 $\times$ 3 symmetric matrix with five of its entries as 1 and four of its entries as 0, we can keep in diagonal either 2 zero and one1 or no zero and three 1 so that the left over zeros and ones are even in number. Hence taking 2 zeros and one1 in diagonal the possible cases are $\frac{3!}{2!}\times \frac{3!}{2!}=9$ and taking 3 ones in diagonal the possible cases are $1\times \frac{3!}{2!}=3$
Total elements A can have = 9 + 3 = 12