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Finding all invertible $A$, a $2\times 2$ matrix that satisfies $A = A^{-1}$ and $A^{-1} = A^T$. Hint: The identity $\cos^2t + \sin^2t = 1$ may be useful.

I have no idea how to start this. Any help would be much appreciated.

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marked as duplicate by rschwieb, Davide Giraudo, Brandon Carter, Thomas, Ron Gordon Feb 11 '13 at 15:10

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What does $A^{-1}1$ mean? Is it simply $A^{-1}$? –  1015 Feb 11 '13 at 12:34
    
@julien Yes, it's A to the power of -1 (i.e. the inverse of A) but I didn't know how to type it properly. –  user61825 Feb 11 '13 at 12:39
    
Edited, you simply needed $\{$ and $\}$ around the $-1$. –  1015 Feb 11 '13 at 12:42
    
Have you checked Orthogonal Matrix ? and Elementary Construction of Them ? –  Inquest Feb 11 '13 at 14:36
    
See here this may be useful [link][1] [1]:math.stackexchange.com/questions/300067/…. –  i.a.m Feb 11 '13 at 14:54
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1 Answer 1

Hint: We have:

  • $A^T=A$ so the matrix is symmetric. So if $A_{2\times 2}=\left( \begin{array}{ccc} a & b \\ c & d \\ \end{array} \right)$ then $c=b$

  • $A^{-1}=A,~~~ A_{2\times 2}$ and if $|A|=ad-bc\neq 0$ so one possibility is $c=0$.

  • If $c\neq 0$ , then $b\neq 0$ and we could have $ad-bc=ad-c^2=-1$ and ...
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+1 nicely done! –  amWhy Feb 11 '13 at 16:12
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