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Math Help - [SOLVED] Matrix A X = C need a lil help kick start

  1. #1
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    [SOLVED] Matrix A X = C need a lil help kick start

    Okay so I've got a question here where I am asked to solve the following matrix equation for X.

    Matrix A
    1 -1 1
    2 3 0
    0 2 -1

    Mattrix X is unknown

    Matrix C

    2 -1 5 7 8
    4 0 -3 0 1
    3 5 -7 2 1

    I have to solve; Matrix A * Matrix X = Matrix C

    How can I find Matrix X? I just need a lil boost to get things rolling.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Well, you should fill in the matrix X with unknowns (make sure you get the dimension right) and expand the product AX. You will get a bunch of linear equations which you can then solve.
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  3. #3
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    Okay so I will find the inverse of A and then multiply it by C to obtain X.

    If this is what I must do to find X I have started and I'm hesitant to continue past this point:

    1 -1 1 : 1 0 0
    2 3 0 : 0 1 0
    0 2 -1 : 0 0 1

    What I'm doing here is reducing the left matrix A to reduced row echelon form in order to find the right matrix A^-1 (inverse of A).

    I encounter a point where I get many many many ratios

    1 -1 1 : 1 0 0
    0 5 -2 : -2 1 0
    0 2 -1 : 0 0 1

    At this point I must multiply row 2 by 1/5 to get a leading 1 to kill the second column but then everything is going to be all screwy and chalk full of ratios.

    Then after it is reduced, I have to multiply the inverse matrix by C to get the unknown matrix X?

    Seems over the top to me, there must be a more efficient way to do this or I'm doing something wrong.
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  4. #4
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    Do you have a problem with using fractions (ratios)? No one said the arithmetic had to be trivial!

    You can save yourself a little work by doing the reduction to the inverse and multiplying that by C at the same time: replace the identity matrix in your first line by C and reduce:
    \begin{bmatrix}1 & -1 & 1 \\2 & 3 & 0\\0 & 2 & -1\end{bmatrix}\begin{bmatrix}2 & -1 & 5 & 7 & 8 \\ 4 & 0 & -3 & 0 & 1 \\ 3 &  5 & -7 & 2 & 1\end{bmatrix}

    Subtract twice the first row from the second row:
    \begin{bmatrix}1 & -1 & 1 \\0 & 5 & -2\\0 & 2 & -1\end{bmatrix}\begin{bmatrix}2 & -1 & 5 & 7 & 8 \\ 0 & 2 & -13 & -14 & -15\\ 3 &  5 & -7 & 2 & 1\end{bmatrix}

    If you really don't want to divide by 5, you could swap the second and third rows (also a legal row operation) and then divide by 2. Since the determinant of A is -1, any fractions will disappear in the end.
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  5. #5
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    Step 1 : Reduce matrix A and obtain the inverse A^-1

    I did this and here is my inverse of A

    row 1: -11/6 1 -1/4
    row 2: -4/6 2/6 3/4
    row 3: -4/3 2/3 -1

    Step 2 : Take the inverse matrix A and multiply by C?

    Step 3: Resulting matrix of A^-1 * C = matrix X???
    Last edited by thekrown; February 13th 2010 at 08:56 AM.
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  6. #6
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    That can't possibly be the inverse of A Multiplying the top row of that, [-11/6 1 -1/4], by the first column of A, [1 2 0], gives -11/6+ 2 which is NOT 1.
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  7. #7
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    Quote Originally Posted by thekrown View Post
    Step 1 : Reduce matrix A and obtain the inverse A^-1

    I did this and here is my inverse of A

    row 1: -11/6 1 -1/4
    row 2: -4/6 2/6 3/4
    row 3: -4/3 2/3 -1

    Step 2 : Take the inverse matrix A and multiply by C?

    Step 3: Resulting matrix of A^-1 * C = matrix X???
    What method of finding the inverse are you using? I get no fractions in the final inverted matrix. The method I like using, is equating your original matrix, with the identity matrix, and every row operation you use on your original matrix (to put it in RRE form), you do to your identity matrix. This will convert the identity matrix into the matrix inverse.

    The first step would go like this:

    \begin{bmatrix} 1 & -1 & 1\\2 & 3 & 0\\0 & 2 & -1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}
    -2R_1+R_3
    \begin{bmatrix} 1 & -1 & 1\\0 & 5 & -2\\0 & 2 & -1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\2 & 1 & 0\\0 & 0 & 1\end{bmatrix}

    To give you a boost, A^{-1} I calculated to be:

    \begin{bmatrix}3 & -1 & 3 \\-2 & 1 & -2\\-4 & 2 & -5\end{bmatrix}

    Now it is up to you to go through your reduction steps, and see if you get the same matrix I did. If not, something went kablam somewhere in your calculations.
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  8. #8
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    I GOT IT!! Thank you all very very much. I've been on this thing for so many hours it's not even funny. Somehow I knew the technique, doubted it worked the second all these ratios started appearing and then my brain started to scramble.

    Anywho, I ended up with the inverted matrix listed above.

    Finally I can do the next step now is to multiply the inverted matrix by C and the resulting matrix will be X? For example, matrix X will also be size 3x5?
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  9. #9
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    Alright guys here are my results.

    First of all, here is original matrix A
    1 -1 1
    2 3 0
    0 2 -1

    Here is inverted matrix A^-1
    3 -1 3
    -2 1 -2
    -4 2 5

    Here is matrix C
    2 -1 5 7 8
    4 0 -3 0 1
    3 5 -7 2 1

    And finally, matrix X
    11 12 -3 27 26
    -6 -8 1 -18 -17
    15 29 -61 -18 -25

    How I got there:

    3*2 + -1*4 + 3*3 = 11
    3*-1 + -1*0 + 3*5 = 12
    etc

    I did this for the entire matrix.

    So, it's been a long journey and I think I understand how to find the unknown matrix X. It would be nice to have someone recheck my work, but not necessary. As long as the method I used to multiply A^-1 and C together is correct, all I need to do is be careful and have patience when finding the inverse with all those ratios. Thank you all.
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  10. #10
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    It looks right. Assuming you did everything right you are good to go.
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