A `3xx3` determinant has entries either `1` or `-1`.
Let `S_(3)=` set of all determinants which contain determinants such that product of elements of any row or any column is `-1` For example `|{:(1,,-1,,1),(1,,1,,-1),(-1,,1,,1):}|`is an element of the set `S_(3)`.
Number of elements of the set `S_(3)=`

Options

A. `10`
B. `16`
C. `12`
D. `18`

Explanation

To solve the problem of finding the number of elements in the set \( S_3 \), we need to analyze the properties of a \( 3 \times 3 \) determinant with entries of either \( 1 \) or \( -1 \), where the product of the elements in any row or column must equal \( -1 \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Each row and each column of the determinant must have a product of \( -1 \). This implies that in each row (or column), there must be an odd number of \( -1 \) entries since the product of an odd number of \( -1 \) values is \( -1 \).

2. **Choosing Entries for the First Two Rows**:
For the first two rows, we can choose any combination of \( 1 \) and \( -1 \) such that there is an odd number of \( -1 \) entries.

- For a row of three entries, the possible combinations that yield an odd product are:
- One \( -1 \) and two \( 1 \)s: There are \( \binom{3}{1} = 3 \) ways to choose which entry is \( -1 \).
- Three \( -1 \)s: There is \( 1 \) way to choose this combination.

Therefore, the total number of valid combinations for each row is:
\[ 3 + 1 = 4 \]
Thus, for the first two rows, the total combinations are:
\[ 4 \times 4 = 16 \]

3. **Determining the Last Row**:
The last row must also ensure that the product of its entries is \( -1 \). The entries in the last row depend on the products of the first two rows. Specifically, the last entry in each column must be chosen to ensure that the product of the column equals \( -1 \).

- If the product of the first two rows in a column is \( 1 \), then the last entry must be \( -1 \).
- If the product of the first two rows in a column is \( -1 \), then the last entry must be \( 1 \).

Since the last row's entries are completely determined by the first two rows, there is only \( 1 \) valid way to fill in the last row for each combination of the first two rows.

4. **Final Calculation**:
Since we have \( 16 \) combinations for the first two rows and \( 1 \) way to fill the last row for each of these combinations, the total number of valid \( 3 \times 3 \) determinants in the set \( S_3 \) is:
\[ 16 \times 1 = 16 \]

Thus, the number of elements in the set \( S_3 \) is \( \boxed{16} \).