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Math Help - What is the inverse of this matrix?

  1. #1
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    What is the inverse of this matrix?

    What is the inverse of:
    [3 7]
    [1 3]

    I've tried working it out, but I keep getting the wrong answer.
    Here's my working out:

    [3 7 1 0]
    [1 3 0 1]

    Divide row 1 by 3
    [1 2.3 0.3 0]
    [1 3 0 1]

    Row 1 minus (1 x row 1)
    [1 2.3 0.3 0]
    [0 1 -0.43 1.43]

    Row 2 divided by 0.7
    [1 2.3 0.3 0]
    [0 1 -0.43 1.43]

    Row 1 minus (2.3 x row 2)
    [1 0 1.289 -3.289]
    [0 1 -0.43 1.43]

    The answer is supposed to be:
    [3/2 -7/2]
    [-1/2 3/2]

    But my answers don't match the ones I got above even when I try and change the decimals to fractions.

    Help would be much appreciated. Sorry if the matrices look messy, I hope you get the idea.
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  2. #2
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    Quote Originally Posted by brumby_3 View Post
    What is the inverse of:
    [3 7]
    [1 3]

    I've tried working it out, but I keep getting the wrong answer.
    Here's my working out:

    [3 7 1 0]
    [1 3 0 1]

    Divide row 1 by 3
    [1 2.3 0.3 0]
    [1 3 0 1]

    Row 1 minus (1 x row 1)
    [1 2.3 0.3 0]
    [0 1 -0.43 1.43]

    Row 2 divided by 0.7
    [1 2.3 0.3 0]
    [0 1 -0.43 1.43]

    Row 1 minus (2.3 x row 2)
    [1 0 1.289 -3.289]
    [0 1 -0.43 1.43]

    The answer is supposed to be:
    [3/2 -7/2]
    [-1/2 3/2]

    But my answers don't match the ones I got above even when I try and change the decimals to fractions.

    Help would be much appreciated. Sorry if the matrices look messy, I hope you get the idea.

    I really cannot follow what you did because it came out too messed up, ...and in fact you shouldn't be asking this: just multiply the original matrix by what you got and if you get the identity matrix you're done, otherwise you're wrong.

    About the decimals: drop them! If you divide by three then write \frac{7}{3}\,\,\,or\,\,\,7\slash 3, and not that horrible thing 2.3 , which is not even a good approximation and it looks childish and,again, pretty ugly. Simple fractions are as numbers as anything else, so befriend them!

    No wonder you got all that mess if you jump between decimal and simple fractions...

    Tonio
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  3. #3
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    Wow, your reply was REALLY rude. You could have been a lot nicer, some people aren't experts at maths, ok? I did the best I could.
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  4. #4
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    Hello, brumby_3!

    With a little thought, we can often avoid fractions.


    Find the inverse of: . \begin{bmatrix} 3&7 \\  1&3 \end{bmatrix}

    We have: . \left[\begin{array}{cc|cc} 3&7 & 1&0 \\ 1&3 & 0&1\end{array}\right]


    \begin{array}{c}R_1-2R_2 \\ \\ \end{array} \left[\begin{array}{cc|cc}1 & 1 & 1 & \text{-}2 \\ 1 & 3 & 0 & 1 \end{array}\right]


    \begin{array}{c} \\ R_2-R_1\end{array} \left[\begin{array}{cc|cc}1 & 1 & 1 & \text{-}2 \\ 0 & 2 & \text{-}1 & 3 \end{array}\right]


    . . . \begin{array}{c} \\ \frac{1}{2}R_2 \end{array} \left[\begin{array}{cc|cc} 1&1&1&\text{-}2 \\ 0 & 1 & \text{-}\frac{1}{2} & \frac{3}{2} \end{array}\right]


    \begin{array}{c}R_1-R_2 \\ \\ \end{array} \left[\begin{array}{cc|cc}1 & 0 & \frac{3}{2} & \text{-}\frac{7}{2} \\ \\[-4mm] 0 & 1 & \text{-}\frac{1}{2} & \frac{3}{2} \end{array}\right]

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  5. #5
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    Quote Originally Posted by brumby_3 View Post
    Wow, your reply was REALLY rude. You could have been a lot nicer, some people aren't experts at maths, ok? I did the best I could.
    I, of course, only tried to stress some arguments I think will help you in the future. In college\university, things will really get tough.

    If you really think my reply was rude (as opposed to you being hypersensitive) then you must send a complain about me to the site's moderators.

    Tonio
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  6. #6
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    There's a difference between not sugar coating things and being rude. In this thread I don't see much sugar but I also don't see anything rude.

    What I do see in fact are members who have given up their valuable time, gratis, to offer good advice and help. Advice I might add that, if followed, will almost certainly lead to improved performance.

    Since the question has been satisfactorily answered, I'm closing the thread to avoid things going off-topic.

    If anyone has anything else to ask or add, s/he can pm it to me.

    Thread closed.
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