Question

Let \( A \) be a square matrix all of whose entries are integers. Then, which one of the following is true?

• A. If \( \text{det} (A) = \pm 1 \), then \( A^{-1} \) exists but all its entries are not necessarily integers.
• B. If \( \text{det} (A) \neq \pm 1 \), then \( A^{-1} \) exists and all its entries are non-integers.
• C. If \( \text{det} (A) = \pm 1 \), then \( A^{-1} \) exists and all its entries are integers.
• D. If \( \text{det} (A) = \pm 1 \), then \( A^{-1} \) need not exist.

Hint:

For matrices with integer entries, if \( \text{det}(A) = \pm 1 \), then the inverse \( A^{-1} \) will also have integer entries.

Answer 1

The Correct Option is C

Step 1: Understanding the condition for the inverse.
For a square matrix \( A \), the inverse \( A^{-1} \) exists if and only if \( \text{det}(A) \neq 0 \). In this case, we are concerned with matrices where the entries of \( A \) are integers, and we are analyzing the behavior of the inverse matrix based on the determinant condition.

Step 2: Matrix inversion when \( \text{det}(A) = \pm 1 \).
We know that the inverse of a matrix \( A \) is given by: \[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \] Where \( \text{adj}(A) \) is the adjugate of \( A \), and its entries are polynomials of the entries of \( A \). When \( \text{det}(A) = \pm 1 \), this implies that the inverse matrix is simply the adjugate matrix, and all the entries in the adjugate matrix will also be integers, because the adjugate is formed from determinants of integer matrices.

Step 3: Conclusion.
Thus, if \( \text{det}(A) = \pm 1 \), then \( A^{-1} \) exists and all its entries are integers. This makes option (3) the correct answer.