# Finding all invertible matrices $A$ where $A = A^{-1}$ and $A^{-1} = A^T$ [duplicate]

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.

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

• What does $A^{-1}1$ mean? Is it simply $A^{-1}$? – Julien 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 $-1$. – Julien 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

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 ...