by Divergent13
 P: 48 Hello everyone the following problem has me completely stumped, I am to find a certain 3x3 matrix D that satisfies the following equation: $$ADA^{-1}$$ = $$\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$ where : $$A = \left(\begin{array}{ccc}1&2&3\\0&1&1\\0&2&1\end{array}\right)$$ $$A^{-1} = \left(\begin{array}{ccc}1&-4&1\\0&-1&1\\0&2&-1\end{array}\right)$$ Heres my reasoning (or lack thereof), I know that $$AA^{-1}$$ will yield the identity matrix I3, however clearly the D im looking for is WITHIN this operation, and by matrix multiplication i cannot use this fact since the order is now completely different. But what I do know is how to find the inverse of A, but what property can I use for finding a 3x3 matrix? You see this would be simpler if they were happening to look for a 3x1 matrix D where I could use row operations in gauss jordan elimination to solve for the particular values, however I did not find any examples of this problem in the book--- where I am given an unknown nxn matrix to find and a certain operation that it must adhere to. I could i solve this one? I have been understanding everything up to this point but i am clearly not understanding some simple rule--- thanks a lot for your help.
 P: 186 lol...its ok :) EDIT: changed to tex $$ADA^{-1}=\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$ $$A^{-1}\left(ADA^{-1}=\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)\right)$$ $$A^{-1}ADA^{-1}=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$ $$DA^{-1}=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$ $$\left(DA^{-1}=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)\right)A$$ $$DA^{-1}A=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)A$$ $$D=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)A$$