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How can the inverse of matrix $A = \left( \begin{smallmatrix} 6&5\\5&4 \end{smallmatrix} \right)$ be $A^{-1} = \left( \begin{smallmatrix} -4&5\\ 5&-6 \end{smallmatrix} \right)$ where $\frac{1}{ad-bc} = \frac{1}{24-25} = \frac{1}{-1}$?

I thought that an inverse to this matrix was not possible, but my textbox and Wolfram Alpha says otherwise. Can someone tell me how this is possible, or if I have misunderstood the formula?

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    $\begingroup$ why did you think it was not possible? $\endgroup$
    – Federica Maggioni
    May 18, 2013 at 11:15
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    $\begingroup$ If you multiply the two matrices, do you not get the identity matrix? What's the problem? $\endgroup$
    – Gerry Myerson
    May 18, 2013 at 11:16
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    $\begingroup$ I'm an idiot. I didn't calculate 1/(-1) correctly $\endgroup$
    – Sam
    May 18, 2013 at 11:18

1 Answer 1

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You can see that this matrix is invertible in couple of ways.

For example:

  1. $|A|=6\cdot4-5\cdot5=-1\neq0$

  2. The rows of $A$ are linearly independent .

  3. The columns of $A$ are linearly independent .

  4. $0$ is not an eigenvalue of $A$ (you can calculate the eigenvalues by finding the characteristic polynomial of $A$)

  5. You have found a matrix $B$ s.t $AB=BA=I$

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  • $\begingroup$ thanks for the great answer $\endgroup$
    – Sam
    May 18, 2013 at 11:24
  • $\begingroup$ @Sam you're welcome! I'm glad you liked it $\endgroup$
    – Belgi
    May 18, 2013 at 11:25

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