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Let's say we have an invertible matrix $A$ and both $A$ and $A^{-1}$ members are whole numbers.

is it always true that $\det A = \det A^{-1}$?

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what is $A^\prime$? The inverse of $A$? This is usually denoted $A^{-1}$. And for invertible matrices you always have $\det(A)= 1/\det(A^{-1})$. –  user20266 Dec 7 '11 at 18:57
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I presume $A'$ is the transpose of $A$. In that case the answer is yes. –  Robert Israel Dec 7 '11 at 19:02
    
uhmm -- yes. But you do not need to assume anything about $A$ in this case (like being invertible, having integer coefficients,...), do you? –  user20266 Dec 7 '11 at 19:09
    
Yeah i ment A' as the invers of A –  Some1 Dec 7 '11 at 19:14
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2 Answers

If $A$ is invertible, then $1 = \operatorname{det} I = \operatorname{det} (AA^{-1}) = \operatorname{det} A \cdot \operatorname{det} A^{-1}$. If both $A$ and $A^{-1}$ have integer entries then their determinant is also an integer. Now, if the product of two integers is $1$ then they have to be both equal to $1$ or $-1$. So, yes, $\operatorname{det} A = \operatorname{det} A^{-1}$.

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HINT $\ $ Multiplicative maps always preserve inverses. If the map $\rm\:d\ne 0\:$ is multiplicative, i.e. $\rm\:d(x\:y) = d(x)\ d(y),\:$ applying $\rm\:d\:$ to $\rm\:a\:a^{-1} = 1\:$ yields $\rm\: d(a)\ d(a^{-1}) = 1\:,\:$ so $\rm\:d(a^{-1}) = d(a)^{-1}\:.$

Note that in a field (or domain) $\rm\:d(1) = 1\:$ since applying $\rm\:d\:$ to $\:1^2 = 1\:$ yields $\rm\:d(1)^2 = d(1),\:$ so $\rm\:d(1) = 1\ or\:\ 0\:.\:$ But if $\rm\:d(1) = 0\:$ then $\rm\:d(x) = d(x\cdot 1) = d(x)\ d(1) = 0,\:$ so $\rm\:d=0\:,\:$ contra hypothesis.

Since $\rm\:\rm\:d = det\:$ is multiplicative, $\rm\:d(a^{-1}) = d(a)^{-1}\:$ so $\rm\:d(a) = d(a^{-1})\ \iff\ d(a) = d(a)^{-1}\:$ $\:\iff\:$ $\rm\:d(a)^2 = 1\:.\:$ But $\rm\:d(a)\:$ is an invertible, and in your case is an integer, so it's $\rm\:\pm1\:,\:$ and $\:(\pm1)^2 = 1\:.\: $ Therefore we conclude that the result holds true if $\rm\:d\:$ takes values in a ring all of whose units are self-inverse, a.k.a. involutary or involutions.

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