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anonymous

  • 5 years ago

another one

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    that last page was making my browser laggy

  3. anonymous
    • 5 years ago
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    sorry:S

  4. anonymous
    • 5 years ago
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    soo?:)

  5. anonymous
    • 5 years ago
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    trying to figure out a clever way to do it so I don't have to do all that multiplication

  6. anonymous
    • 5 years ago
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    well I know that A^2n = I and A^2n+1 = A from my matrix calculator

  7. anonymous
    • 5 years ago
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    so the solution and the steps is A^2n = I and A^2n+1 = A only??

  8. anonymous
    • 5 years ago
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    yeah, but you are going to need to show how you got that

  9. anonymous
    • 5 years ago
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    yeah actually:S:S sorry

  10. anonymous
    • 5 years ago
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    okay: for the muliplication on the diagonal we get: (1/2)^2 + (-1/2)^2 + (-1/2)^2 + (-1/2)^2 = 1

  11. anonymous
    • 5 years ago
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    for not on the diagonal we get: 1/2*-1/2 + 1/2*-1/2 + 1/2*1/2 + 1/2*1/2 = 0

  12. anonymous
    • 5 years ago
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    see I did part a but the problem in part B

  13. anonymous
    • 5 years ago
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    So A^2 = identity matrix then A^3 = A^2 * A = Identity * A = A

  14. anonymous
    • 5 years ago
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    well then for A^2m it's just (A^2)^n which is just multiplying an Identity matrix a bunch of times

  15. anonymous
    • 5 years ago
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    for A^2n+1 = A^2n * A = (A^2)^n * A = A

  16. anonymous
    • 5 years ago
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    thats all?

  17. anonymous
    • 5 years ago
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    yep A^2n = (A^2)^n = Id^n = Id A^2n+1 = (A^2)^n * A= Id * A = A

  18. anonymous
    • 5 years ago
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    thanks:)

  19. anonymous
    • 5 years ago
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    find conditions on a and b such that the following system of linear equations has a no solution b unique solution or c infinitely solution x+y+3z=2 x+2y+4z=3 x+3y+az=b

  20. anonymous
    • 5 years ago
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    hmm

  21. anonymous
    • 5 years ago
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    sorry Iam making u tired :)

  22. anonymous
    • 5 years ago
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    not sure how to effectively do that one.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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