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Math Help - Could someone please help me with these complicated matrix operations!?

  1. #1
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    Could someone please help me with these complicated matrix operations!?

    Hello, I am doing an independent study on matrix operations and made a few problems up so someone can show me how to break them down and solve. I am really having a hard time understanding them and I figure that the best way to learn them would be from the work and each individual step. If someone has the time, please help me by breaking each of these four problems down for me, so I know how to do them. I appreciate anyone who is willing to help.

    IMPORTANT: Each section of brackets, for example, as in question #1, is to be read as the first row [2 3] or second row [3 5] of one matrix. I had to ask because I wrote it according to some poorly written problems and didn't understand what it meant.

    1. [2 3] [3 5]

    2. [6 8] [5 7]

    Find the matrix X:

    3. [6 5] [4 -2] * X = [18 49] [-20 6]

    4. [3 4] [-3 -2] * X = [10 -10] [-8 8]

    Last edited by bobbyboy1111; August 22nd 2009 at 10:16 PM.
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    Quote Originally Posted by bobbyboy1111 View Post
    Hello, I am doing an independent study on matrix operations and made a few problems up so someone can show me how to break them down and solve. I am really having a hard time understanding them and I figure that the best way to learn them would be from the work and each individual step. If someone has the time, please help me by breaking each of these four problems down for me, so I know how to do them. I appreciate anyone who is willing to help.

    1. [2 3] [3 5]

    2. [6 8] [5 7]

    Find the matrix X:

    3. [6 5] [4 -2] * X = [18 49] [-20 6]

    4. [3 4] [-3 -2] * X = [10 -10] [-8 8]

    first, you need to know that one cannot multiply a (1 x 2) matrix with another (1 x 2) matrix.
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    Quote Originally Posted by bobbyboy1111 View Post
    1. [2 3] [3 5]
    What does that notation mean?
    What is to be done with it once it is defined?
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    Quote Originally Posted by skeeter View Post
    first, you need to know that one cannot multiply a (1 x 2) matrix with another (1 x 2) matrix.
    I wrote these according to some algebra problems I had last year and they could be multiplied, so these are more than correct...
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    Quote Originally Posted by Plato View Post
    What does that notation mean?
    What is to be done with it once it is defined?

    The notation means multiply.
    Find the inverse.
    Last edited by bobbyboy1111; August 22nd 2009 at 10:43 AM.
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    Quote Originally Posted by bobbyboy1111 View Post
    The notation means multiply.
    Find the inverse.
    That makes no sense at all! If the "notation means multiply", then what are you to find the inverse of? As skeeter told you before, you cannot multiply a (1 by 2) matrix by a (1 by 2) matrix.
    You can multiply a (1 by 2) matrix by a (2 by 1) matrix. Did you mean \left[\begin{array}{cc}2 & 3\end{array}\right]\left[\begin{array}{c}2 \\ 5\end{array}\right]

    Or do you mean the dot product of two vectors?

    Or, since you now mention inverses, do you mean to find the inverse of the matrix \left[\begin{array}{cc}2 & 3 \\ 2 & 5\end{array}\right]?
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    Quote Originally Posted by HallsofIvy View Post
    That makes no sense at all! If the "notation means multiply", then what are you to find the inverse of? As skeeter told you before, you cannot multiply a (1 by 2) matrix by a (1 by 2) matrix.
    You can multiply a (1 by 2) matrix by a (2 by 1) matrix. Did you mean \left[\begin{array}{cc}2 & 3\end{array}\right]\left[\begin{array}{c}2 \\ 5\end{array}\right]

    Or do you mean the dot product of two vectors?

    Or, since you now mention inverses, do you mean to find the inverse of the matrix \left[\begin{array}{cc}2 & 3 \\ 2 & 5\end{array}\right]?
    I appologize, I was thinking of two other problems.. No, they are just next to each other as written above and the inverse needs to be found. No other notation, no multiplication.
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    Does the notation 1. [2 3] [3 5] mean \left( {\begin{array}{*{20}c}   2 & 3  \\3 & 5  \\\end{array} } \right)?
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    Quote Originally Posted by Plato View Post
    Does the notation 1. [2 3] [3 5] mean \left( {\begin{array}{*{20}c} 2 & 3 \\3 & 5 \\\end{array} } \right)?
    I don't think so.. Do you think I should ask my old teacher?
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    Quote Originally Posted by bobbyboy1111 View Post
    I don't think so.. Do you think I should ask my old teacher?
    By all means do.
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    Quote Originally Posted by Plato View Post
    By all means do.

    O.k., I will get back late tonight or early tomorrow.
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    Quote Originally Posted by HallsofIvy View Post
    That makes no sense at all!

    Or, since you now mention inverses, do you mean to find the inverse of the matrix \left[\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right]?

    My old teacher emailed me back and said it is to be read as a complete matrix, like you have written above, in all of the problems I made up.
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    Quote Originally Posted by bobbyboy1111 View Post
    My old teacher emailed me back and said it is to be read as a complete matrix, like you have written above, in all of the problems I made up.
    Good! Now, I know two methods of finding inverse matrices, one using the determinant and "cofactors", the other "row operations".
    With your matrix, \left[\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right], the determinant is just \left|\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right|= 2(5)- (3)(3)= 1. If you label the numbers is a matrix " a_{ij}", the cofactor of a_{ij} is the determinant you get by removing the ith row and jth column, multiplied by (-1)^{i+j}. For this simple 2 by 2 matrix, the cofactor of "2", in the first row, first column, is just (-1)^{1+1}(5)= 5, the cofactor of the "3" in first row, second column, is (-1)^{1+2}(3)= -3, the cofactor of the "3" in second row, first column is (-1)^{2+ 1}(3)= -3 and the cofactor of the "5" in second row, second column, is (-1)^{2+2}(2)= 2. The matrix you get by replacing each number by its cofactor is \left[\begin{array}{cc}5 & -3 \\ -3 & 2\end{array}\right]. The inverse matrix is just that with each number divided by the determinant. Since, here, the determinant was 1, the inverse of \left[\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right] is just \left[\begin{array}{cc}5 & -3 \\ -3 & 2\end{array}\right].
    You should try multiplying \left[\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right]\left[\begin{array}{cc}5 & -3 \\ -3 & 2\end{array}\right] and \left[\begin{array}{cc}5 & -3 \\ -3 & 2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right] to see that both do, in fact, give the inverse matrix.

    For simple, 2 by 2, matrices (especially if the determinant is 1!) that probably is the simplest method. For larger matrices it is much simpler to use "row operations". The three kinds of row operations are (1) swap two rows, (2) multiply every number in a row by the same number, and (3) add a multiple of one rwo to another.

    Set up the given matrix and the identity matrix side by side. Here:
    \left[\begin{array}{cc}2 & 3 \\ 3 & 5\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]

    Now perform row operations to change the original matrix to the identity matrix- at the same time performing the row operation to the matrix on the right as well. When you have converted the original matrix to the identity matrix, you will have converted the identity matrix to the inverse of the original matrix.

    It is best to work one column at a time, getting "1" in the appropriate position (called the "pivot"), then getting the "0"s you need in that column. Here, the "first column, first row position, is "2" so I can get a "1" in that position by multiplying the entire first row by 1/2:
    \left[\begin{array}{cc}1 & \frac{3}{2} \\ 3 & 5\end{array}\right]\left[\begin{array}{cc}\frac{1}{2} & 0 \\ 0 & 1\end{array}\right]
    After I have a "1" in that position, I can get a "0" in the second row, first column by subtracting 3 times that new first row from the second row:
    \left[\begin{array}{cc}1 & \frac{3}{2} \\ 0 & \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}\frac{1}{2} & 0 \\ -\frac{3}{2} & 1\end{array}\right]
    The first column is now complete so I turn to the second column. I need a "1" where there is now " \frac{1}{2}" so I need to multiply the second row by 2:
    \left[\begin{array}{cc}1 & \frac{3}{2} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{2} & 0 \\ 0 & 2\end{array}\right]

    Finally, get "0" in the first row, second column, by subtracting 3/2 times the new second row from the first row:
    \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}5 & -3 \\ -3 & 2\end{array}\right]
    exactly the matrix we got before.
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    What about the ones for find matrix X?
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    Quote Originally Posted by bobbyboy1111 View Post
    What about the ones for find matrix X?
    [A] \cdot [X] = [{B}]

    [A]^{-1} \cdot [A] \cdot [X] = [A]^{-1} \cdot [{B}]

    [{I}] \cdot [X] = [A]^{-1} \cdot [{B}]

    [X] = [A]^{-1} \cdot [{B}]
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