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Math Help - what am i doing wrong? finding the inverse of the matrix

  1. #1
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    what am i doing wrong? finding the inverse of the matrix

    the question is find the inverse the matrix.

    [3 1][1 0]
    [4 2][0 1]

    in Calc Chat (step by step) they multiplied 1/2(R2) first then did -R2 + R1 and so forth.

    I thought I was suppose to multiply 1/3(R1) first to get 1 on R1 and so forth. I am so confused. I know how to set it up but I still dont get why I could have just multiplied 1/3(R1). ??? In Gaus-elimination are you suppose to make sure the coefficent is 1 on the first row??

    thank you
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  2. #2
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    Re: what am i doing wrong? finding the inverse of the matrix

    Hello, MariaBlood777!

    \text{Find the inverse of: }\:A \;=\;\begin{bmatrix}3&1\\4&2\end{bmatrix}

    \left[\begin{array}{cc|cc}3&1&1&0 \\4&2&0&1\end{array}\right]

    in Calc Chat (step by step) they multiplied 1/2(R2) first then did -R2 + R1 and so forth.
    I can see why they did that, but it's a very strange opening move.

    I thought I was suppose to multiply 1/3(R1) first to get 1 on R1 and so forth.
    That is correct, but there are other ways to get 1 on R1.

    We can do it "your way", but it will introduce fractions.
    If you can handle them, great!

    \text{We have: }\:\left[\begin{array}{cc|cc}3&1&1&0 \\ 4&2&0&1 \end{array}\right]

    . . . . \begin{array}{c}\frac{1}{3}R_1 \\ \frac{1}{4}R_2 \end{array}\:\left[\begin{array}{cc|cc} 1&\frac{1}{3} & \frac{1}{3}&0 \\ 1 & \frac{1}{2} & 0 & \frac{1}{4}\end{array}\right]

    . \begin{array}{c}\\ R_2-R_1 \end{array}\:\left[\begin{array}{cc|cc} 1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & \frac{1}{6} & \text{-}\frac{1}{3} & \frac{1}{4} \end{array}\right]

    . . . . \begin{array}{c}\\6R_2 \end{array}\:\left[\begin{array}{cc|cc}1 & \frac{1}{3} & \frac{1}{3} & 0 \\ 0 & 1 & \text{-}2 & \frac{3}{2}\end{array}\right]

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


    Therefore: . A^{-1} \;=\;\begin{bmatrix}1 & \text{-}\frac{1}{2} \\ \text{-}2 & \frac{3}{2} \end{bmatrix}

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