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Homework Help: Linear Algebra: Proof

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    |A| + |B| = |C|
    where
    A=
    [a_11 a_12 a_13]
    [a_21 a_22 a_23]
    [a_31 a_32 a_33]
    B=
    [b_11 b_12 b_13]
    [b_21 b_22 b_23]
    [b_31 b_32 b_33]
    C=
    [(a_11 + b_11) a_12 a_13]
    [(a_21 + b_21) a_22 a_23]
    [(a_31 + b_31) a_32 a_33]

    2. Relevant equations

    ?

    3. The attempt at a solution

    First I tried taking the determinants of all 3 matrices but it got like WAY too messy and I didnt see anything close to them being equal (since i got a lot of terms that involved a_ij x b_ij and I didnt get them on the other ones.

    Then I researched a little and came across that
    A=
    [a_11 a_12 a_13] = [a_11 a_12 a_13] + [ 0 a_12 a_13] + [0 a_12 a_13]
    [a_21 a_22 a_23] [0 a_22 a_23] [a_21 a_22 a_23] [0 a_22 a_23]
    [a_31 a_32 a_33] [0 a_32 a_33] [0 a_32 a_33] [a_31 a_32 a_33]


    I think it goes somewhere around there but I keep getting lost and I dont know how to get around it.
     
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 29, 2010 #2

    Mark44

    Staff: Mentor

    If A and B are 3 x 3 matrices, then A + B is also a 3 x 3 matrix whose typical entry is cij = aij + bij.

    For example, c11 = a11 + b11, which is what you have, but c12 = a12 + b12. Instead, you have a12 without b12.

    The matrix you show for C can't be the sum of A and B.
    Determinants don't enter into a problem about the sum of matrices. Since you seem to think that they are involved in this problem, it might be that you haven't given the exact problem statement.
     
  4. Sep 29, 2010 #3
    wait let me say cause I didnt specify, sorry for the big mistake I meant to say
    Proove that:

    |A| + |B| = |C|

    my super bad sorry!
    Im talking about determinants all the time, my bad used [] instead of ||
     
  5. Sep 30, 2010 #4

    Mark44

    Staff: Mentor

    What you're trying to prove isn't true, so you're going to have a tough time proving it.

    Here's a counter example.
    [tex]A = \left[\begin{array}{c c c}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right][/tex]
    [tex]B = \left[\begin{array}{c c c}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right][/tex]
    [tex]C = \left[\begin{array}{c c c}4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right][/tex]
    I picked these matrices to make calculating the determinants easier.
    |A| = 1, |B| = 27, and |C| = 4
    So |A| + |B| [itex]\neq[/itex] |C|

    I don't think you copied the problem correctly, particularly matrix B.
     
  6. Sep 30, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    Actually, |A|+ |B|= |C| makes no sense if you give no more information about A, B, and C. Surely it is not true for any three matrices- for A and B fixed, |A|+ |B| is fixed and is not equal to |C| for all matrices |C|.

    I suspect you mean "If A, B, and C are 3 by 3 matrices such that A+ B= C, then |A|+ |B|= |C|." That now makes sense (before, you had no "if" part) but, as Mark44 said, still is not true.
     
  7. Sep 30, 2010 #6
    yes! OMG you guys are correct. . .I cant believe I overlooked it

    the actual problem says the following:


    |A| + |B| = |C|
    where
    A=
    [a_11 a_12 a_13]
    [a_21 a_22 a_23]
    [a_31 a_32 a_33]
    B=
    [b_11 a_12 a_13]
    [b_21 a_22 a_23]
    [b_31 a_32 a_33]
    C=
    [(a_11 + b_11) a_12 a_13]
    [(a_21 + b_21) a_22 a_23]
    [(a_31 + b_31) a_32 a_33]


    so the way I was going around was correct :) (I think. . .Im gonna start working it out but I believe i was on the right track )
     
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