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Math Help - Inverse of a matrix?

  1. #1
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    Inverse of a matrix?

    Hi, I'm having a problem finding the inverse of a matrix in that I have no idea where I am going wrong. I am given two simultaneous equations and told to find the inverse of A and therefore determine the value of x.

     2x_1 + 3x_2 = 1\  and\  x_1 + 7x_2 = 6 . I have put this in the form Ax = b as asked by the question.

    \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right] = \left[\begin{array}{c}1\\6\end{array}\right].

    I know that if I multiply both sides by the inverse I get x = A^{-1}b. To find the inverse from here A^{-1} = \frac{1}{det(A)}\cdot   A_{cof}T (where the T is for transposed).

    So A = \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \ \ A_{cof} = \left[\begin{array}{cc}2&-1\\-3&7\end{array}\right] \ \ A_{cof}T = \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]

    I calculated the det(A) as being 11 (14 -3) so A^{-1} = \frac{1}{11} \cdot \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]

    If something hasn't already gone wrong then here is where it doesn't make sense from me. Just to make sure that it was correct I decided to multiply A by the inverse I just got and to see if I got the identity matrix. I didn't. Can anyone help me in finding where I have gone wrong. I can do the rest from there

    Thanks
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Beard View Post
    Hi, I'm having a problem finding the inverse of a matrix in that I have no idea where I am going wrong. I am given two simultaneous equations and told to find the inverse of A and therefore determine the value of x.

     2x_1 + 3x_2 = 1\  and\  x_1 + 7x_2 = 6 . I have put this in the form Ax = b as asked by the question.

    \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right] = \left[\begin{array}{c}1\\6\end{array}\right].

    I know that if I multiply both sides by the inverse I get x = A^{-1}b. To find the inverse from here A^{-1} = \frac{1}{det(A)}\cdot   A_{cof}T (where the T is for transposed).

    So A = \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \ \ A_{cof} = \left[\begin{array}{cc}2&-1\\-3&7\end{array}\right] \ \ A_{cof}T = \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]

    I calculated the det(A) as being 11 (14 -3) so A^{-1} = \frac{1}{11} \cdot \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]

    If something hasn't already gone wrong then here is where it doesn't make sense from me. Just to make sure that it was correct I decided to multiply A by the inverse I just got and to see if I got the identity matrix. I didn't. Can anyone help me in finding where I have gone wrong. I can do the rest from there

    Thanks
    Your error is in finding A_{cof}T, which should be::

      \left[\begin{array}{cc}7&-3\\-1&2\end{array}\right]
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  3. #3
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    Hello, Beard!

    Your inverse matrix is wrong . . .


    given: . \begin{array}{ccc} 2x + 3y &=& 1 \\ x + 7y &=& 6 \end{array}

    I have put this in the form Ax = b as asked by the question.

    . . \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x \\ y\end{array}\right] \:=\: \left[\begin{array}{c}1\\6\end{array}\right]


    So: . A^{-1} \:=\: \frac{1}{11} \cdot \left[\begin{array}{cc}2&\text{-}3\\\text{-}1&7\end{array}\right] . . . . no

    There is a "formula" for the inverse of a 2-by-2 matrix.


    We have: . \left[\begin{array}{cc} a&b \\ c&d \end{array}\right]

    [1] Switch the elements on the main diagonal: . \left[\begin{array}{cc}{\color{red}d}&b\\c&{\color{red}a  }\end{array}\right]

    [2] Change the signs on the other diagonal: . \left[\begin{array}{cc}d & {\color{red}\text{- }}\!b \\ {\color{red}\text{- }}\!c & a\end{array}\right]

    [3] Divide by the determinant: . {\color{red}\frac{1}{ad-bc}}\,\left[\begin{array}{cc} d & \text{- }\!b \\ \text{- }\!c & a\end{array}\right]


    Therefore: . \left[\begin{array}{cc}a&b\\c&d\end{array}\right]^{-1} \;=\;\frac{1}{ad-bc}\,\left[\begin{array}{cc}d&\text{- }\!b \\ \text{- }\!c & a \end{array}\right]

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  4. #4
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    I know can see how I went wrong, the examples I was trying to follow had ones for a and c looking at sorobans method so I was unable to distinguish.

    Thanks
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