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Is A^2 equivalent to AxA or all the elements of A are squared?

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    I was just wondering if say A is a 2x2 matrix. Is A^2 equivalent to AxA or all the elements of A are squared?

    2. Relevant equations
    let A and B be 2x2 matrices. is the following true?
    (A-B)(A+B) = A^2 - B^2


    3. The attempt at a solution
     
  2. jcsd
  3. Feb 8, 2010 #2

    statdad

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    Homework Helper

    Re: matrices

    For 1: Raising a square matrix to a positive integer power is defined exactly as it is for numbers: [tex] a^2 [/tex] is short-hand for the product of [tex] a [/tex] with itself.

    For 2: The rule [tex] (A-B)(A+B) = A^2 - B^2 [/tex] is true for numbers because multiplication of reals is commutative. FOIL out the left side, and look for the spot where that property is used for real numbers in order to get to [tex] A^2 - B^2 [/tex]. Is the corresponding statement true for matrix multiplication?
     
  4. Feb 8, 2010 #3
    Re: matrices

    does that mean A^2 = AxA?
    erm.. yes?
     
  5. Feb 8, 2010 #4
    Re: matrices

    Yes, A2=AxA
     
  6. Feb 8, 2010 #5

    HallsofIvy

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    Re: matrices

    Yes, A^2= A*A.

    No, (A- B)(A+ B) is not equal to A^2- B^2. It is equal to A^2+ AB- BA- B^2 but the AB and BA do not cancel because matrix multiplication is not commutative.
     
  7. Feb 8, 2010 #6
    Re: matrices

    ahh I see :)
    I cancelled out AB-BA ignoring the fact that matrix multiplication is not commutative
    thanks!

    then I suppose (A-I)(A^2 + A + I) = A^3 - I is true?

    attempt : expand LHS

    = A^3 + A^2 + A - IA^2 - IA - I^2
    = A^3 + A^2 + A - A^2 - A - I
    = A^3 - I (since I^2 would be I right?)
     
    Last edited: Feb 8, 2010
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