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Homework Help: Determinant Products

  1. Aug 4, 2014 #1
    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. jcsd
  3. Aug 4, 2014 #2


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    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.
  4. Aug 4, 2014 #3
    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...
    Last edited: Aug 4, 2014
  5. Aug 4, 2014 #4
    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: Aug 4, 2014
  6. Aug 4, 2014 #5


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    You could state that the multiplications and additions are done in Zp and the various operations work the same in Zp as in R.
  7. Aug 4, 2014 #6
    I can say this since Zp is in R? I'm confused as to the exact relationship between the two.
  8. Aug 4, 2014 #7
    You can say that because [itex]R\in\mathbb{M}_{2\times 2}[/itex]!
  9. Aug 4, 2014 #8
    So the ring is an element of the 2*2 matrices?
  10. Aug 4, 2014 #9
    Right on! So your proof applies to it as well.
  11. Aug 4, 2014 #10
    Ah, that makes sense now! Thank you.
  12. Aug 4, 2014 #11


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    Don't you mean ##\subseteq## rather than ##\in##?
  13. Aug 5, 2014 #12
    Oh well, you're right. Sorry!
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