# Determinant Products

1. Aug 4, 2014

### Justabeginner

1. The problem statement, all variables and given/known data
Let R be the ring of all 2*2 matrices over Zp , a prime. Show that for x, y contained in R, det(xy) = det(x) det(y).

2. Relevant equations

3. The attempt at a solution
The det(xy)≠0, therefore the equality can be true. However, I am not sure how to prove that the equality is true without using a value, say det(xy)=1?

2. Aug 4, 2014

### Fredrik

Staff Emeritus
Since the matrices are only 2×2, it's not too much work to just let x and y be two arbitrary matrices and then explicitly compute xy, det x, det y and finally (det x)(det y) and det xy.

3. Aug 4, 2014

### Justabeginner

So if X = (a b c d)
and Y = (e f g h)
XY = (ae + bg af + bh ce + dg cf + dh)

I may say that
Det(Y) = eh - fg
Det(XY) = [(ae + bg)(cf + dh)] - [(af + bh)(ce + dg)]
(aecf + bgcf + aedh + bgdh) - (afce + bhce + afdg + bhdg)
bgcf + aedh - bhce - afdg

Is the above reasoning correct? I thought it should be more complicated since it involves R being the ring of all 2*2 matrices over Zp, a prime, and I haven't considered that in my argument...

Last edited: Aug 4, 2014
4. Aug 4, 2014

### patbuzz

Hi!
Everything seems all right. In particular, your proof is applicable in the ring $R$ since $R\in\mathbb{M}_{2\times 2}$. Binet-Cauchy formula states that for any square matrix $A,B$ of the same order, $$det(AB)=det(A) det(B) = det(BA)$$ Hence, that property of determinants has nothing to do with the primality of the entries of the matrices.

Last edited: Aug 4, 2014
5. Aug 4, 2014

### HallsofIvy

Staff Emeritus
You could state that the multiplications and additions are done in Zp and the various operations work the same in Zp as in R.

6. Aug 4, 2014

### Justabeginner

I can say this since Zp is in R? I'm confused as to the exact relationship between the two.

7. Aug 4, 2014

### patbuzz

You can say that because $R\in\mathbb{M}_{2\times 2}$!

8. Aug 4, 2014

### Justabeginner

So the ring is an element of the 2*2 matrices?

9. Aug 4, 2014

### patbuzz

Right on! So your proof applies to it as well.

10. Aug 4, 2014

### Justabeginner

Ah, that makes sense now! Thank you.

11. Aug 4, 2014

### Fredrik

Staff Emeritus
Don't you mean $\subseteq$ rather than $\in$?

12. Aug 5, 2014

### patbuzz

Oh well, you're right. Sorry!