Proving det(xy) = det(x) det(y) for R Matrices Over Zp

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In summary, we have shown that for any two matrices x and y contained in the ring R of all 2*2 matrices over Zp, their determinants multiplied together is equal to the determinant of their product. This result holds true regardless of the values of the elements in the matrices and is due to the Binet-Cauchy formula. This property applies to R because it is itself an element of the 2*2 matrices.
  • #1
Justabeginner
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Homework Statement


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).


Homework Equations





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?
 
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  • #2
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
Fredrik said:
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.

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(X) = ad - bc
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
(ad - bc)(eh) - (ad - bc)(fg)

Det(X)Det(Y)= (ad)(eh) - (bc)(eh) - (ad)(fg) + (bc)(fg)
(ad- bc)(eh) - (ad - bc)(fg)

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...
 
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  • #4
Hi!
Everything seems all right. In particular, your proof is applicable in the ring [itex]R[/itex] since [itex]R\in\mathbb{M}_{2\times 2}[/itex]. Binet-Cauchy formula states that for any square matrix [itex]A,B[/itex] of the same order, [tex]det(AB)=det(A) det(B) = det(BA)[/tex] Hence, that property of determinants has nothing to do with the primality of the entries of the matrices.
 
Last edited:
  • #5
Justabeginner said:
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(X) = ad - bc
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
(ad - bc)(eh) - (ad - bc)(fg)

Det(X)Det(Y)= (ad)(eh) - (bc)(eh) - (ad)(fg) + (bc)(fg)
(ad- bc)(eh) - (ad - bc)(fg)

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...
You could state that the multiplications and additions are done in Zp and the various operations work the same in Zp as in R.
 
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  • #6
HallsofIvy said:
You could state that the multiplications and additions are done in Zp and the various operations work the same in Zp as in R.

I can say this since Zp is in R? I'm confused as to the exact relationship between the two.
 
  • #7
You can say that because [itex]R\in\mathbb{M}_{2\times 2}[/itex]!
 
  • #8
patbuzz said:
You can say that because [itex]R\in\mathbb{M}_{2\times 2}[/itex]!

So the ring is an element of the 2*2 matrices?
 
  • #9
Right on! So your proof applies to it as well.
 
  • #10
patbuzz said:
Right on! So your proof applies to it as well.

Ah, that makes sense now! Thank you.
 
  • #11
patbuzz said:
[itex]R\in\mathbb{M}_{2\times 2}[/itex].
Don't you mean ##\subseteq## rather than ##\in##?
 
  • #12
Oh well, you're right. Sorry!
 

1. What is the significance of proving det(xy) = det(x) det(y) for R Matrices over Zp?

The significance of this proof lies in its application in fields such as cryptography, coding theory, and linear algebra. It allows for efficient computation of determinants for R matrices over Zp, which have a wide range of practical uses.

2. What is the definition of a R matrix over Zp?

A R matrix over Zp is a matrix whose elements are integers modulo p, where p is a prime number. These matrices are used to represent and manipulate data in various mathematical applications.

3. How does one go about proving det(xy) = det(x) det(y) for R matrices over Zp?

The proof involves using basic properties of determinants, such as linearity and the product rule, along with the fact that integers modulo p form a field. By carefully manipulating the terms, one can show that the equality holds for all R matrices over Zp.

4. What are some potential challenges in proving det(xy) = det(x) det(y) for R matrices over Zp?

One potential challenge is dealing with the complex algebraic operations involved in working with R matrices over Zp. Another challenge may be in finding a clear and concise way to present the proof, as it can involve multiple steps and equations.

5. Are there any alternative approaches to proving det(xy) = det(x) det(y) for R matrices over Zp?

Yes, there are multiple approaches to proving this equality. Some involve using properties of group theory, while others use more advanced concepts such as abstract algebra and linear algebra. Ultimately, the most efficient approach may vary depending on the specific context and purpose of the proof.

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