Finding a 3x3 Matrix D that Satisfies a Given Equation

[Divergent13]

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 I am 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.

 

[Spectre5]

Multiply both sides on the left by A inverse, then multiply both sides on the right by A...then D is on the left and you can expand out the other to find what D is.

 

[Divergent13]

Wait what do you mean by expanding out the other? What does both sides on the left mean? Like A and D? Thanks.

 

[Divergent13]

btw i really apologize for the stupid thread title... i was ctually testing out my TeX format and accidentally posted with a wrong name--- id change it if i could but i cannot!

 

[Spectre5]

lol...its ok :)

EDIT: changed to tex

$$ADA^{-1}=\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$

$$A^{-1}\left(ADA^{-1}=\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)\right)$$

$$A^{-1}ADA^{-1}=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$

$$DA^{-1}=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)$$

$$\left(DA^{-1}=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)\right)A$$

$$DA^{-1}A=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)A$$

$$D=A^{-1}\left(\begin{array}{ccc}1&0&0\\1&0&0\\1&0&0\end{array}\right)A$$