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Math Help - inverse matrices

  1. #1
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    Question inverse matrices

    How would I prove the inverse of this matrice is as stated


    1 1 1
    1 2 3 is inverse of
    1 3 6

    3-3 1
    -3 5-2
    1-2 1

    with working out if possible
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  2. #2
    TD!
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    Show that the product is equal to the unit matrix.
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  3. #3
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    Quote Originally Posted by TD! View Post
    Show that the product is equal to the unit matrix.
    Both ways
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  4. #4
    TD!
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    If you're referring to having to check AB = I and BA = I, that's not true.
    It's sufficient to show only one, if AB = I, then BA = I as well.
    Although matrix multiplication doesn't commute in general, it does for A and its inverse.
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  5. #5
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    Hello, Greg!

    Since you are studying Inverses, I assume you know how to multiply.
    What's stopping you?


    Code:
                | 1 1 1 |                   |  3 -3  1 |
    Prove that: | 1 2 3 | is the inverse of | -3  5 -2 |
                | 1 3 6 |                   |  1 -2  1 |
    Code:
        | 1 1 1 | |  3 -3  1 |
        | 1 2 3 |.| -3  5 -2 |
        | 1 3 6 | |  1 -2  1 |
    
        | 1(3) + 1(-3) + 1(1)   1(-3) + 1(5) + 1(-2)   1(1) + 1(-2) + 1(1) |
      = | 1(3) + 2(-3) + 3(1)   1(-3) + 2(5) + 3(-2)   1(1) + 2(-2) + 3(1) |
        | 1(3) + 3(-3) + 6(1)   1(-3) + 3(5) + 6(-2)   1(1) + 3(-2) + 6(1) |
    
        | 3 - 3 + 1   -3 +  5 - 2    1 - 2 + 1 |
      = | 3 - 6 + 3   -3 + 10 - 6    1 - 4 + 3 |
        | 3 - 9 + 6   -3 + 15 - 12   1 - 6 + 6 |
    
        | 1  0  0 |
      = | 0  1  0 |
        | 0  0  1 |
    . . . . . . . . .
    Use the slide-bar to see the rest of line 2.

    Edit: corrected the typo!
    Last edited by Soroban; October 14th 2006 at 05:21 PM.
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  6. #6
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    I disagree.

    | 1 0 1 |
    | 0 1 0 |
    | 0 0 1 |

    Is not the identity matrix. A way to check this would be to augment the original Matrix with it's identity matrix and row reduce it to see what the inverse is.

    That is, in this case,

    M := Matrix([[3, -3, 1, 1, 0, 0], [-3, 5, -2, 0, 1, 0], [1, -2, 1, 0, 0, -1]])

    (Using Maple notation). In which case, once that is row reduced, you get:

    Matrix([[1, 0, 0, 1, 1, -1], [0, 1, 0, 1, 2, -3], [0, 0, 1, 1, 3, -6]])

    The ABOVE matrix is the inverse matrix (in augmented columns), that is:

    Matrix([[1, 1, -1], [1, 2, -3], [1, 3, -6]])
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  7. #7
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    Thanks for your help

    I know this stuff is really easy, but I have not studied maths much and am doing a distance learning course, I really appreciate your help. It makes it a lot easier to understand when I see the working out. Thank you again
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  8. #8
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    Quote Originally Posted by AfterShock View Post
    I disagree.

    | 1 0 1 |
    | 0 1 0 |
    | 0 0 1 |

    Is not the identity matrix. A way to check this would be to augment the original Matrix with it's identity matrix and row reduce it to see what the inverse is.

    That is, in this case,

    M := Matrix([[3, -3, 1, 1, 0, 0], [-3, 5, -2, 0, 1, 0], [1, -2, 1, 0, 0, -1]])

    (Using Maple notation). In which case, once that is row reduced, you get:

    Matrix([[1, 0, 0, 1, 1, -1], [0, 1, 0, 1, 2, -3], [0, 0, 1, 1, 3, -6]])

    The ABOVE matrix is the inverse matrix (in augmented columns), that is:

    Matrix([[1, 1, -1], [1, 2, -3], [1, 3, -6]])
    I'm sorry, but my calculator checks that the original pair are indeed inverse matrices, both by multiplication and by taking the inverse. Your above matrix is NOT an inverse for the one with the top row [3, -3, 1].

    -Dan
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  9. #9
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    Indeed, that is because:

    Matrix([[3, -3, 1, 1, 0, 0], [-3, 5, -2, 0, 1, 0], [1, -2, 1, 0, 0, -1]])


    When I was row reducing that, I had a -1, where it should have been positive. My answer is correct if you simply change the sign of the last column.
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