If $AD−BC=I_2$, does $\begin{pmatrix} A & B \\ C & D\end{pmatrix}^{-1}$ = $\begin{pmatrix} D & -B \\ -C & A\end{pmatrix}$?

Question

Let A,B,C,D be 2×2 matrices. If $AD-BC$ = $I_2$ is it true that $\begin{pmatrix} A & B \\ C & D\end{pmatrix}^{-1}$ = $\begin{pmatrix} D & -B \\ -C & A\end{pmatrix}$ ?"

My approach: Let $M=\begin{pmatrix} A & B \\ C & D\end{pmatrix}$ and $N=\begin{pmatrix} D & -B \\ -C & A\end{pmatrix}$. I then check if $MN$ equals $I_4$, but find that $MN=\begin{pmatrix} AD - BC & -AB + BA \\ CD - DC & -CB + DA \end{pmatrix}$. Does this mean the proposition is false? (I assume so because BC does not generally equal CB, and similarly, AB does not generally equal BA.)

Welcome to MSE! As you are a new contributor to the site. I suggest you read some basic information about writing mathematics at this site, see, e.g. How can I format mathematics, There should be universal LaTeX/MathJax guide for sites supporting it, and MathJax basic tutorial and quick reference — Afntu, Jul 11 '24 at 10:36
If $A,\ldots,D$ are themselves square matrices, all you apparently need to present is a single explicit example where $-AB+BA\ne 0$ holds and also $AD-BC=I_2$ — Hagen von Eitzen, Jul 11 '24 at 10:52
@lulu sorry my bad, there were some descriptive issues in my post content，it might be clearer right now — Yukai HUA, Jul 11 '24 at 11:00
Seconding what hagen said --- the top right matrix ($-AB+BA$) would have to be all zeros (in order for $MN$ to equal $I_4$). [Same with the bottom left ($CD-DC$)]. This is the case where $AB=BA$ and $CD=DC$. We already know that $AD-BC=I_2$. Whatever condition could make $-CB+DA=I_2$ would finish it off. So, we would say that the proposition is not true in general, but true in the special case we've started to line out here. (Mainly that $A,B,C,D$ would have to commute). — ness, Jul 11 '24 at 11:40

If $AD−BC=I_2$, does $\begin{pmatrix} A & B \\ C & D\end{pmatrix}^{-1}$ = $\begin{pmatrix} D & -B \\ -C & A\end{pmatrix}$?

0 Answers0