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Expert-verified Found in: Page 165 ### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384 # Question: 18. Suppose that all the entries in A are integers and $${\bf{det}}\,A = {\bf{1}}$$. Explain why all the entries in $${A^{ - {\bf{1}}}}$$ are integers.

Here, all the entries in $${A^{ - 1}}$$ are integers. This is because all entries in the adjugate matrix of A are integers.

See the step by step solution

## Step 1: Describe the given statement

Given, all entries in A are integers and $$\det A = 1$$.

## Step 2: Use the formula for cofactors

The cofactor of A is:

$${C_{ij}} = {\left( { - 1} \right)^{i + j}}{A_{ij}}$$

Since $${A_{ij}}$$ is the sum of the product of entries in A, $${A_{ij}}$$ is an integer. Thus $${C_{ij}}$$ is an integer.

Hence, all the cofactors of A are integers. This implies that all the entries in the adjugate matrix of A are integers.

## Step 3: Use Theorem 8

By Theorem 8,

$$\begin{array}{c}{A^{ - 1}} = \frac{1}{{\det A}}{\rm{adj}}\,A\\ = \frac{1}{1}{\rm{adj}}\,A\\{A^{ - 1}} = {\rm{adj}}\,A\end{array}$$

Hence, we conclude that all entries in $${A^{ - 1}}$$ are integers. ### Want to see more solutions like these? 