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Math Help - Evaluate

  1. #1
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    Evaluate

    Another one I'm not sure on is this one:

    Evaluate
    {1 5}{1 2}
    {2 3}{0 1}

    Possible solutions:
    {1 10}
    {0 3 }
    -------
    {1 10}
    {0 7}
    -------
    {1 7}
    {5 7}
    -------
    {1 7}
    {2 7}
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  2. #2
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    Quote Originally Posted by gretchen View Post
    Another one I'm not sure on is this one:

    Evaluate
    {1 5}{1 2}
    {2 3}{0 1}
    first row first column (1,5)(1,0)' = 1.1+5.0=1
    first row second column = (1,5)(2,1)'=1.2+5.1=7

    second row first column = (2,3)(1,0)'=2.1+3.0=2
    second row second colum = (2,3)(2,1)'=2.2+3.1=7

    so the matrix product is:

    [ 1 7 ]
    [ 2 7 ]

    RonL
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  3. #3
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    Thanks Ron, you have helped me so much, if only you knew. Because you show me how you get the answer, I am getting the other problems like them correct.

    I have two other types of evaluations if you could help me out:

    First one is similar to the one before but it takes a 3x3 matrix:

    Q: Evaluate
    {3 2 1}{2}
    {3 1 0}{2}
    {0 -2 1}{2}

    Solutions:
    {8}
    {6}
    {1}
    -----
    {12}
    {8}
    {-2}
    -----
    {12}
    {8}
    {1}
    -----
    {12}
    {6}
    {-2}


    Second one is finding the inverse:

    Q: Find A[-1 power] where A=
    {2 4}
    {2 5}

    Solutions:
    {5 -2}
    {-4 2}
    -----
    {5 -4}
    {-2 2}
    -----
    {1 -2}
    {-1 2.5}
    -----
    {2.5 -2}
    {-1 1}
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  4. #4
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    Quote Originally Posted by gretchen View Post
    Thanks Ron, you have helped me so much, if only you knew. Because you show me how you get the answer, I am getting the other problems like them correct.

    I have two other types of evaluations if you could help me out:

    First one is similar to the one before but it takes a 3x3 matrix:

    Q: Evaluate
    {3 2 1}{2}
    {3 1 0}{2}
    {0 -2 1}{2}
    First element is [3, 2, 1][2, 2, 2]' = 3.2 + 2.2 + 1.2 = 12
    Second element is [3, 1, 0][2, 2, 2]' = 3.2 + 1.2 + 0.2 = 8
    Third element is [0, -2, 1][2, 2, 2]' = 0.2 + (-2).2 + 1.2 = -2

    so the answer is:

    [12]
    [ 8]
    [-2]

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by gretchen View Post
    Second one is finding the inverse:

    Q: Find A[-1 power] where A=
    {2 4}
    {2 5}

    Solutions:
    {5 -2}
    {-4 2}
    -----
    {5 -4}
    {-2 2}
    -----
    {1 -2}
    {-1 2.5}
    -----
    {2.5 -2}
    {-1 1}
    Given that you are given a list of options the easiest way to do this is
    to multiply each of the candidates by the original matrix.

    Start with the first row first column.


    Case 1:
    [2, 4][5,-4]' = 2.5-4.4 = -6 != 1, so not this one.

    Case 2:
    [2, 4][5,-2]' = 2.5-4.2 = -2 != 1, so not this one.

    Case 3:
    [2, 4][1,-1]' = 2.1-4.1 = -2 != 1, so not this one.

    Case 4:
    [2, 4][2.5,-1]' = 2.(2.5)-4.1 = 1 != 1, so this is the one, now just
    check the other elements of the product are the elements of the
    identity matrix.

    RonL
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