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Q18E

Expert-verifiedFound in: Page 165

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.

Given, all entries in *A* are integers and \(\det A = 1\).

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.

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.

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