Show that $A \in GL_2(\mathbb Z_2)$ if and only if $\det(A) \neq 0$ and $A^{-1} = \det(A)^{-1} B$.

Question

Where $B = \begin{bmatrix} [d] & [-b] \\ [-c] & [a] \end{bmatrix}$

The way I solved this problem is That I first got all the elements of $M_2(\mathbb Z_2)$ and then considered the elements that have non-zero determinant so this is the elements of $GL_2(\mathbb Z_2)$, however this approach has depended on the fact that knowing $GL_2(\mathbb Z_2)$ elements , however I was wondering is there a more direct approach of solving this problem without relying on knowing the elements $GL_2(\mathbb Z_2)$?

Stephen Montgomery-Smith · Accepted Answer · 2015-02-22T20:45:49.390

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You can show that $AB = BA = \det(A) I$ just by multiplying them out. Hence if $\det(A) \ne 0$, you have constructed the inverse using the formula given in your title.

To show the converse, use the fact that $\det(UV) = \det(U) \det(V)$. (It's true for matrices of integers, so it must also be true for matrices of $\mathbb Z_2$.) Hence $\det(A^{-1}) \det(A) = 1$.

edited Feb 22 '15 at 20:45

answered Feb 22 '15 at 04:13

Stephen Montgomery-Smith

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however I have a question we need to prove that det(A) $\neq $ $0$ for the $\rightarrow$ implication. – Illustionist Feb 22 '15 at 19:51
also I don't see how is AB = det(A)I can you please elaborate more ? – Illustionist Feb 22 '15 at 19:55
For example AB we have in first entry a1b1 = [a1b1 + a2b3] a1b2 = [a1b2 + a2b4] etc, however when we multiply det(A)I we get something completely different. Could maybe explain a little bit more on those points please ? – Illustionist Feb 22 '15 at 19:59
When I multiply out $AB$, I get $\begin{bmatrix} [ad-bc] & 0 \ 0 & [ad-bc] \end{bmatrix}$. – Stephen Montgomery-Smith Feb 22 '15 at 20:27
Also, if there is something that multiplied by $\det(A)$ is $1$, then we must have $\det(A) \ne 0$. If $\det(A) = 0$, we must have $\det(A) \det(A^{-1}) = 0$, which is a contradiction. – Stephen Montgomery-Smith Feb 22 '15 at 20:30
Never mind I understand now.. – Illustionist Feb 22 '15 at 20:36
I see okay I understand everything fully now thank you for your answer !! – Illustionist Feb 22 '15 at 20:37
I guess their is also 1 step missing is we have to also check AB = det(A)I = BA and then we now have constructed the inverse. – Illustionist Feb 22 '15 at 20:39
Yes, you do have to show that both $AB$ and $BA$ are $\det(A) I$. Internally I was remembering that for any finite dimensional matrix over a field it has a left inverse if and only if it has a right inverse. But that is non-trivial, and it is easier to also multiply out $BA$. I'll modify my answer. – Stephen Montgomery-Smith Feb 22 '15 at 20:44
Stephen Montgomery-Smith could you explain a bit for the converse I don't understand the converse... ? – Illustionist Feb 22 '15 at 23:08
how come if we know that determinant $det(A^-1)det(A) = 1 = det(A^-1 * A) = det(A*A^-1)$ – Illustionist Feb 22 '15 at 23:09
Then this will will imply that is in $Gl_2(Z_3)$ ? – Illustionist Feb 22 '15 at 23:10
The formula $\det(UV) = \det(U) \det(V)$ is true for $n\times n$ matrices over any commutative ring. http://math.stackexchange.com/questions/1159738/showing-that-detab-det-a-det-b-with-the-following-identity/1159781#1159781 – Stephen Montgomery-Smith Feb 22 '15 at 23:36
Yes, this proof will show that for $2\times2$ matrices over any commutative ring, a matrix is invertible if and only if its determinant is invertible. – Stephen Montgomery-Smith Feb 22 '15 at 23:38
The result is also true for $n \times n$ matrices. In place of the matrix $B$, use the Adjugate matrix http://en.wikipedia.org/wiki/Adjugate_matrix – Stephen Montgomery-Smith Feb 22 '15 at 23:40
I see I thought in general it wouldn't be true but yeh this makes sense thanks again! – Illustionist Feb 23 '15 at 00:04

Show that $A \in GL_2(\mathbb Z_2)$ if and only if $\det(A) \neq 0$ and $A^{-1} = \det(A)^{-1} B$.

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