Question

Let \(A\) be a matrix each of whose entries are integers. Show that each of the following conditions implies the other. 1\. \(A\) is invertible and \(A^{-1}\) has integer entries. 2\. \(\operatorname{det} A=1\) or -1

Short answer

Both conditions are equivalent: one implies the other.

Step-by-step solution

Understand the problem

We are given a matrix \(A\) with integer entries and need to show an equivalence between \(A\) being invertible with integer entries in \(A^{-1}\) and \(\operatorname{det} A = \pm 1\). This involves proving two implications: (1 implies 2) and (2 implies 1).

Prove 1 implies 2

Assume \(A\) is invertible and \(A^{-1}\) has integer entries. The determinant of a matrix is a product of its eigenvalues (for an invertible matrix, none are zero), and the determinant of the matrix is the reciprocal of the determinant of its inverse. Since \(A\) and \(A^{-1}\) have integer entries, it follows that \(\operatorname{det}(A) \cdot \operatorname{det}(A^{-1}) = 1\). Therefore, \(\operatorname{det}(A)\) must be \(\pm 1\) to keep the product as an integer because it has an inverse in the integers.

Prove 2 implies 1

Assume \(\operatorname{det} A = 1\) or \(-1\). If \(\operatorname{det}(A)\) is \(\pm 1\), \(A\) is invertible. The inverse \(A^{-1} = \frac{1}{\operatorname{det}(A)} \cdot \operatorname{adj}(A)\) will also have integer entries, because \(\operatorname{adj}(A)\), the adjugate of \(A\), has integer entries. Since we are dividing by \(\pm 1\), the entries in \(A^{-1}\) remain integers.

Conclusion

We have shown that both conditions are equivalent. If \(A\) is invertible and \(A^{-1}\) has integer entries, this forces \(\operatorname{det} A = \pm 1\). Conversely, if \(\operatorname{det} A = \pm 1\), then \(A\) is invertible, and \(A^{-1}\) will have integer entries.

Key concepts

Matrix Invertibility

When we talk about a matrix being invertible, we mean that there is another matrix that, when multiplied with the original matrix, results in the identity matrix. In mathematical terms, a matrix \( A \) is invertible if there exists a matrix \( A^{-1} \) such that \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix with ones on the diagonal and zeros elsewhere.
Understanding invertibility is crucial because not all matrices have inverses. The ability to invert a matrix typically depends on the determinant of the matrix (which we'll discuss further). An invertible matrix is also referred to as a "nonsingular" or "non-degenerate" matrix.

• In practice, an invertible matrix can crucially reverse the transformation it represents.
• Keep in mind, for a matrix to be invertible, it needs to be square, meaning it has the same number of rows and columns.

This concept appears often in solving systems of linear equations, finding linear transformations, and modeling various practical problems.

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. Not only does it help in determining matrix invertibility, but it also shows up in various other areas of linear algebra. Intuitively, the determinant provides information about the volume scaling factor of the linear transformation described by the matrix.
For a matrix \( A \), the determinant is denoted as \( \operatorname{det}(A) \). Its value is crucial:

• If \( \operatorname{det}(A) = 0 \), the matrix is not invertible, meaning it doesn't have an inverse.
• If \( \operatorname{det}(A) eq 0 \), the matrix is invertible, and you can compute its inverse.

The determinant defines whether the matrix's rows (or columns) are linearly independent. In layman's terms, the determinant can help figure out if a matrix has a unique solution or not, particularly in solving linear systems. In our specific problem, if \( \operatorname{det}(A) = \pm 1 \), special properties come into play concerning integer entries in the inverse matrix. This highlights an elegant aspect of working with integer matrices.

Integer Entries in Matrix Inverse

For a matrix with integer entries, having its inverse also contain integer entries is a very interesting case. Normally, when you invert a matrix, the entries of \( A^{-1} \) might not end up as integers. However, there is a clear criterion for when this will indeed occur.
If the determinant of an integer matrix \( A \) is \( \pm 1 \), then not only is the matrix invertible, but also the entries of its inverse \( A^{-1} \) remain integers.
This stands out because for non-special determinants (other than \( \pm 1 \)), the inverse matrix usually has fractional or decimal entries. Hence, integer matrices with determinants of \( \pm 1 \) form an elegant subset of matrices where operations can be performed without leaving the realm of integers. These matrices are useful in computational scenarios where maintaining integer precision is crucial and round-off errors from fractions or decimals need to be avoided. Further, in cryptographic applications and modular arithmetic, handling integers is often paramount for both integrity and simplicity.