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I need to find the area of the image of a circle centred at the origin with radius 3 under the transformation:

$ \begin{pmatrix} 3 & 0\\ 0 & \frac{1}{3} \end{pmatrix} $

The image is the ellipse $ \frac{x^2}{81}+y^2=1$. It would appear that it has the same area as the original circle i.e. $9\pi$. Is this because the matrix has some special property such as being its own inverse?

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  • $\begingroup$ This matrix is not its own inverse. $\endgroup$ May 26 '15 at 11:00
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    $\begingroup$ It's because the matrix has the property that its determinant is 1. By the way your equation doesn't match the matrix, perhaps the equation should be $x^2/9 + 9y^2=1$. $\endgroup$
    – KalEl
    May 26 '15 at 11:01
  • $\begingroup$ Think geometrically. You've stretched the $x$ axis and shrunk the $y$ axis. $\endgroup$ May 26 '15 at 11:02
  • $\begingroup$ Also by the way so you are clear - the matrix is not its own inverse. $\endgroup$
    – KalEl
    May 26 '15 at 11:07
  • $\begingroup$ Yes, thank you for pointing out that it's not its own inverse. That was a big mistake. $\endgroup$ May 26 '15 at 11:09
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Yes, this matrix has a special property, namely its determinant is 1.

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It is a known formula that the area enclosed in the ellipse with semi-axes $a$ and $b$ is $\pi ab$, as may be seen from the orthogonal affinity that transforms a circle into an ellipse.

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