# 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

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!