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Identity Matrices

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data
    0vzUmzv.jpg

    2. Relevant equations
    ImA = AIn = A.
    (A−1)−1 = A
    (AB)−1 = B−1A−1

    3. The attempt at a solution
    Determinates:
    Det(A) = 3 – 0 = 3
    Det (2A+BT) = 4 – 8 = -2

    Matrices
    B^T = 2 -2
    0 -5

    (2A + B^T)^-1 = -8 -4
    -8 -2

    So I've kinda figured out those sides, but I'm having real issues finding the identity matrix for this question. I'm not really sure exactly how to find it. Then for the second question, I just have no idea what it's asking me to do. Thanks for any help.
     
  2. jcsd
  3. Apr 24, 2013 #2

    chiro

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    Hey Siann122 and welcome to the forums.

    The identity matrix for most purposes is just an nxn matrix with 1's in the diagonal position and 0's everywhere else.

    The second question is asking to prove a statement that LHS = RHS. One hint for this would be to use that fact that A * A^(-1) = I and use other properties of matrices to get your answer.
     
  4. Apr 24, 2013 #3
    So the 2I would be a matrix of: 2 0
    2 2

    Because it would have 1 on the diagonals and 0 in the corner as it's a 2*2 matrix?
     
  5. Apr 24, 2013 #4
    No.
    ##
    \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I##
    So ##2I = \left[ \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right]##


    Really, I should be saying ##I_n## (##I_2## in this case), but generally the dimension of the identity matrix is suppressed and you make it as big as whatever it needs to be; it's usually obvious.
     
  6. Apr 24, 2013 #5
    Alright so using the identity matrix I've come up with this result:

    (2A + B^T)^-1 = -

    2 -1
    -2 -1/2

    (A-1 * B^T + 2I) =

    (2 2/3) (-4)
    (1 1/3) (-4 1/3)


    Det(A^-1 * B^T + 2I) = - 44/9 – ( - 16/3) = 4/9

    (A^-1 * B^T + 2I)^-1 =

    6 -9
    3 -9 3/4


    But then (A^-1 * B^T + 2I)^-1 =/= (2A + B^T)^-1 , so what am I doing wrong?

    PS: How did you do the matrix formatting :P
     
  7. Apr 24, 2013 #6

    Dick

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    I wouldn't worry too much about the matrix formatting at this point. I write a matrix like A as [[1,2],[2,3]]. It's quick and it gets the point across. You can worry about the formatting later. Can you show more of what you are doing? Before you say (2A+B^T)^(-1)=[[2,-1],[-2,-1/2]], can you say what you got for (2A+B^T)?
     
  8. Apr 24, 2013 #7
    Here's my full working out.

    A =
    1 2
    2 3
    Det(A) = 3 – 0 = 3

    A-1 =
    1/3 2/3
    2/3 1

    2A+B^T =
    4 2
    4 1
     
  9. Apr 24, 2013 #8

    Dick

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    A^(-1) is wrong and Det(A) is wrong. 2A+B^T looks ok. Can you maybe show a little more working? Why do you think Det(A)=3?
     
  10. Apr 24, 2013 #9
    Sorry, Det(A) should be -1 as (3x1)-(2x2), which would make A^-1 wrong.

    So A^-1 =
    -1 -2
    -2 -3

    Which means A^-1 * B^T =
    (-1)(2)+(-2)(0) (-1)(-2)+(-2)(-5)
    (-2)(2)+(-3)(0) (-2)(-2)+(-3)(-5)
    = -2 12
    -4 19

    That means (A-1 * BT + 2I) =
    (0) (12)
    (-4) (21)

    Det(A-1 * BT + 2I) = - 0 – -48 = 48

    (A-1 * BT + 2I)-1 =
    0 1/4
    -1/12 7/16

    This is still wrong though...
     
  11. Apr 24, 2013 #10

    Dick

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    Now Det(A) is good. But A^(-1) is still wrong. You can check this by multiplying A*A^(-1) and you won't get I. You don't just divide A or A^T by Det(A). Can you show your working there?
     
  12. Apr 24, 2013 #11
    Yep so 1/-1 = -1, and then A^-1 =

    (-1)(1) (-1)(2)
    (-1)(2) (-1)(3)

    As it's only a scalar amount that is being multiplied in.
     
  13. Apr 24, 2013 #12

    Dick

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    Ok, we are getting down to where your problem lies. To get the inverse you divide the transpose of the cofactor matrix of A by Det(A). The cofactor matrix of [[a,b],[c,d]] is [[d,-c],[-b,a]]. Transpose that and divide by Det(A). And go back and review how to find inverses! And as I said before you can easily check whether you have the correct inverse by multiplying A by A^(-1). Do you get I?
     
  14. Apr 24, 2013 #13
    Ah okay, so
    A-1 =
    -3 2
    2 -1 ??
     
  15. Apr 24, 2013 #14
    I edited it while you were posting so you must have missed it, to reiterate it should be:

    A-1 =
    -3 2
    2 -1 ??
     
  16. Apr 24, 2013 #15

    Dick

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    That's more like it. You can check it yourself by doing the multiplication. I recommend that. So continue from there.
     
  17. Apr 24, 2013 #16
    A-1 * BT =
    (-3)(2)+(2)(0) (-3)(-2)+(2)(-5)
    (2)(2)+(-1)(0) (2)(-2)+(-1)(-5)
    = -6+0 6+(-10)
    (-4)+0 (-4)+(5)
    =-6 -4
    4 1

    (A-1 * BT + 2I) =
    (-4) (-4)
    (4) (3)

    Det(A-1 * BT + 2I) = -12 - -16 = 4
    (A-1 * BT + 2I)-1 =
    3/4 -1
    4 -1


    Is this correct?
     
  18. Apr 24, 2013 #17

    Dick

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    Almost. If (A-1 * BT + 2I)=[[-4,-4],[4,3]] (which I agree with) then did you really get the right inverse finally? Check it. I don't think you did. Try checking yourself more. I'm signing off soon so I can't keep this up much longer. But I think you are getting the hang of it.
     
  19. Apr 24, 2013 #18
    Det(A-1 * BT + 2I) = -12 - -16 = 4
    (A-1 * BT + 2I)-1 =
    3/4 1
    -4 -1

    It's this last bit, the -1 should have been 1, and the 4 should have been -4 (I think).
     
  20. Apr 24, 2013 #19

    Dick

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    (I think)!? You should know. You have to learn how to correct yourself at some point. This would be a good time to start. I'll give the hint that it's not correct. And I think you are perfectly capable of doing it correctly if you do it more carefully.
     
  21. Apr 24, 2013 #20
    Haha, basic arithmetic fail, totally overlooked that 1/4 * 4 = 1, so it should be:

    (A-1 * BT + 2I)-1 =
    3/4 1
    -1 -1
     
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