4

I know that the units of 2 by 2 matrices with integer entries must have a determinant of 1 or -1, and I have proved that if the determinant is zero then the matrix is not a unit, however I am wondering how you would go about proving that matrices with determinants other than 1 and -1 are not units?

Jenny
  • 161
  • 1
  • 1
  • 5

2 Answers2

3

An element $\;r\;$ in a ring is a unit if there exists another element $\;x\;$ there s.t. $\;rx=1\;$ .

If $\;A,B \;$ are square integer matrices ,then

$$AB=1\implies \det A=\frac1{\det B}\;$$

But the rightmost number is not an integer if $\;\det B\neq\pm 1\;$ ...

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
1

Given a matrix of integers $$\begin{pmatrix}a & b \\ c & d \end{pmatrix} $$ it's inverse is $$ \frac{1}{ad-bc} \begin{pmatrix}d & -b \\ -c & a \end{pmatrix} .$$ This will be a matrix of integers if and only if $\frac{1}{ad-bc} \in \mathbb Z$.

user1337
  • 24,381
  • 2
    Is it possible for the determinant to divide all of the entries in the matrix. Like if the determinant is 1/2 and all of the entries are even? – Jenny Nov 30 '13 at 19:42
  • 1
    @Jenny if $a=a' \Delta,b=b' \Delta,c=c' \Delta,d=d' \Delta$ where $a',b',c',d'$ are integers and $\Delta =\det A$, you must have $a'd'-b'c'=\frac{1}{\Delta}$. Since $\Delta$ is an integer it has to be $\pm 1$. – user1337 Nov 30 '13 at 19:49
  • Why do we get ad-bc=(ad-bc)^2 if we know that (ad-bc) divides a,b,c,d? – Jenny Nov 30 '13 at 20:11
  • @Jenny With the notations above we get $ad-bc=(a'd'-b'c') \Delta^2 $ rather than $\Delta^2$. – user1337 Nov 30 '13 at 20:14
  • Ignore my previous comment, it was in response to a comment that was deleted. Why don't we say -b=b'detA and -c=c'detA? – Jenny Nov 30 '13 at 20:21
  • Those minus signs are insignificant since we could use the integers $b''=-b',c''=-c'$ instead and return to the previous case. – user1337 Nov 30 '13 at 20:25
  • a' b' c' and d' are the integers you get when you multiply each entry by the determinants? Is that correct? – Jenny Nov 30 '13 at 20:28
  • @Jenny I believe you meant divide – user1337 Nov 30 '13 at 20:41