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Algebraic manipulation of identity matrix

  1. Jun 9, 2005 #1
    I was doing a few practice questions and was a bit confused about how the solution manual manipulated the I identity matrix. For example,

    Just say you had the following

    I - 2A = B

    where A and B are 2x2 matrices. B is given, but we need to find A.

    Therefore, shouldn't it be

    -2A = B - I
    A = -(1/2) * (B - I)?

    Because the book did it like this

    2A = I - B
    A = (1/2) * (1 - B)

    Can anyone tell me why? Because both statements do not seem to be equivalent. (unless I'm missing something)

    Thanks in advance.
  2. jcsd
  3. Jun 9, 2005 #2
    I think you may have had one of those brain lapses that I hate so much. The statements are equivalent, as the negative sign will distribute.

    -(1/2)*(B-I) = (1/2)*-(B-I) = (1/2)*(I-B)
  4. Jun 9, 2005 #3
    Argh, thanks a bunch. So it's just basic algebraic manipulations correct? No special things I have to do to bring I to the other side?
  5. Jun 9, 2005 #4
    Nope, nothing special. Matrix addition and scalar multiplication satisfy all of the properties of regular addition and multiplication (commutativity, associativity, distributivity, etc.), so basic manipulations will suffice.
  6. Jun 9, 2005 #5
    Thanks a lot!
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