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Math Help - Inverting matrix with unknown.

  1. #1
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    Inverting matrix with unknown.

    Hi all,

    Quick question.

    I'm just unsure of how to handle an unknown in a matrix when talking about finding the inverse.

    Usually I simply augment (A|I) and reduce it to RREF and read the inverse off the identity matrix.

    This is the matrix however I have to do it on -

     \begin{array}{cc}-2&1\\1&x\end{array}

    Obviously I'm not (in any way I can think of) able to reduce the x to a zero, so how do I transform this into reduced row echelon form, and hence get the inverse?

    Cheers.
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  2. #2
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    Hi Peleus!

    Quote Originally Posted by Peleus View Post
    Hi all,

    Quick question.

    I'm just unsure of how to handle an unknown in a matrix when talking about finding the inverse.

    Usually I simply augment (A|I) and reduce it to RREF and read the inverse off the identity matrix.

    This is the matrix however I have to do it on -

     \begin{array}{cc}-2&1\\1&x\end{array}

    Obviously I'm not (in any way I can think of) able to reduce the x to a zero, so how do I transform this into reduced row echelon form, and hence get the inverse?

    Cheers.
    This is a 2x2 matrix and there is a formula for this problem

    Invertible matrix - Wikipedia, the free encyclopedia
    (title: Inversion of 22 matrices )


    But you don't want to use this, right?

    Yours
    Rapha
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  3. #3
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    I've only got one question on it, so it will get me by for this, but as always the saying goes give a man a fish and eat for a day, teach a man.....


    Is there any other way of doing it?
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  4. #4
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    Quote Originally Posted by Peleus View Post
    Hi all,

    Quick question.

    I'm just unsure of how to handle an unknown in a matrix when talking about finding the inverse.

    Usually I simply augment (A|I) and reduce it to RREF and read the inverse off the identity matrix.

    This is the matrix however I have to do it on -

     \begin{array}{cc}-2&1\\1&x\end{array}

    Obviously I'm not (in any way I can think of) able to reduce the x to a zero, so how do I transform this into reduced row echelon form, and hence get the inverse?

    Cheers.
    Swap the rows:

    \left( \begin{array}{cc}<br />
1 & x \\<br />
-2 & 1 \end{array}\right)


    R2 \rightarrow R2 + 2 R1:

    \left( \begin{array}{cc}<br />
1 & x \\<br />
0 & 1 + 2x\end{array}\right)


    R2 \rightarrow \frac{1}{1 + 2x} R2 \left( \text{provided } 1 + 2x \neq 0 \Rightarrow x \neq - \frac{1}{2} \right):

    \left( \begin{array}{cc}<br />
1 & x \\<br />
0 & 1\end{array}\right)


    R1 \rightarrow R1 - x R2:

    \left( \begin{array}{cc}<br />
1 & 0 \\<br />
0 & 1 \end{array}\right)


    Note that x \neq - \frac{1}{2} is consistent with the requirement that the determinant is not equal to zero.
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