Let A be a square matrix all of whose entries are integers. Then which one of the following is true? (1) If `d e t A""=+-1,""t h e n""A^(1)` exists but all its entries are not necessarily integers (2) If `d e t A!=""+-1,""t h e n""A^(1)` exists and all its entries are non-integers (3) If `d e t A""=+-1,""t h e n""A^(1)` exists and all its entries are integers (4) If `d e t A""=+-1,""t h e n""A^(1)` need not exist

Options

A. If `det A ne 1,` then `A^(-1)` exists and all its entries are non-integers
B. If `det A = pm 1. ` then `A^(1)` then `A^(-1)` exist and all its entries are integers
C. If `det A = pm 1, ` then `A^(-1)` need not exist
D. If `det A = pm 1, ` then ` A^(-1)` exists but all its entries are not
necessarily integers

Explanation

To solve the problem, we need to analyze the properties of the inverse of a square matrix \( A \) with integer entries based on its determinant.

### Step-by-Step Solution:

1. **Understanding the Inverse of a Matrix**:
The inverse of a matrix \( A \), denoted as \( A^{-1} \), exists if and only if the determinant of \( A \) is non-zero. The formula for the inverse is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
where \( \text{adj}(A) \) is the adjugate of \( A \).

2. **Case When \( \text{det}(A) = \pm 1 \)**:
- If \( \text{det}(A) = 1 \) or \( \text{det}(A) = -1 \), then \( A^{-1} \) exists.
- Since both \( \text{det}(A) \) and \( \text{adj}(A) \) are integers (because \( A \) has integer entries), \( A^{-1} \) will also have integer entries because:
\[ A^{-1} = \frac{1}{\pm 1} \cdot \text{adj}(A) = \text{adj}(A) \]
- Therefore, if \( \text{det}(A) = \pm 1 \), all entries of \( A^{-1} \) will be integers.

3. **Case When \( \text{det}(A) \neq \pm 1 \)**:
- If \( \text{det}(A) \) is any integer other than \( \pm 1 \), \( A^{-1} \) still exists, but the entries of \( A^{-1} \) will not necessarily be integers. This is because \( \frac{1}{\text{det}(A)} \) will not be an integer if \( \text{det}(A) \) is any integer other than \( \pm 1 \).

4. **Evaluating the Options**:
- **Option 1**: If \( \text{det}(A) = \pm 1 \), then \( A^{-1} \) exists but all its entries are not necessarily integers. (Incorrect, as shown above).
- **Option 2**: If \( \text{det}(A) \neq \pm 1 \), then \( A^{-1} \) exists and all its entries are non-integers. (Not necessarily true, as \( A^{-1} \) can have integer entries if \( \text{det}(A) \) is a divisor of the entries of \( \text{adj}(A) \)).
- **Option 3**: If \( \text{det}(A) = \pm 1 \), then \( A^{-1} \) exists and all its entries are integers. (Correct).
- **Option 4**: If \( \text{det}(A) = \pm 1 \), then \( A^{-1} \) need not exist. (Incorrect, as \( A^{-1} \) does exist).

### Conclusion:
The correct statement is **Option 3**: If \( \text{det}(A) = \pm 1 \), then \( A^{-1} \) exists and all its entries are integers.