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I know that over a field $F$, a matrix is invertible if and only if its determinant is nonzero. And I understand why this is true, at least in the case where the field is $\mathbb{R}$.

But I do not understand how to actually compute the inverse over a finite field. In this question, there is a suggestion to use the formula involving the adjugate matrix, that works over $\mathbb{R}$. Here, another user suggests row reduction and provides a helpful step-by-step walkthrough, but the modulus is prime so everything is invertible.

He writes:

For extra credit review the algorithm, and observe that, all the operations and laws of algebra that the algorithm depends on, hold in all fields.

But I can't see why this is true. For instance, what if the determinant of the matrix (which it seems we usually end up dividing by when we take the inverse) has no multiplicative inverse in $F$?

Community
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user79141
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1 Answers1

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By definition, every nonzero element in a field has a multiplicative inverse. When you're working over a non-field like $\mathbb Z_6$, then your worries come true, and you can have two matrices with non-zero determinant whose product is the zero matrix. Can you give some $2\times 2$ examples?

Ted Shifrin
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  • Thanks very much, I'll think about examples, but now I understand what confused me---I didn't really know the definition of a field, I thought all the $\mathbb{Z}_n$ qualified. – user79141 May 23 '13 at 18:12