Talk:Block matrix

I have just added the majority of the current page content, so the previous discussions were outdated, since they referred to a considerably different page.Thiagovscoelho (talk) 15:33, 20 March 2024 (UTC) [reply ]

I believe that the statement "This is also true when ${\displaystyle AB=BA}${\displaystyle AB=BA}, ${\displaystyle AC=CA}${\displaystyle AC=CA}, or ⁠${\displaystyle BD=DB}${\displaystyle BD=DB}⁠." is false.

If ${\displaystyle AC=CA}${\displaystyle AC=CA}, then ${\displaystyle \det {\begin{bmatrix}A&B\\C&D\end{bmatrix}}=\det(AD-CB).}${\displaystyle \det {\begin{bmatrix}A&B\\C&D\end{bmatrix}}=\det(AD-CB).} Note the change in order of ${\displaystyle C}${\displaystyle C} and ${\displaystyle B}${\displaystyle B} (we have ${\displaystyle CB}${\displaystyle CB} instead of ${\displaystyle BC}${\displaystyle BC}). Similarly, if ${\displaystyle BD=DB}${\displaystyle BD=DB}, then ${\displaystyle AD}${\displaystyle AD} should be replaced with ${\displaystyle DA}${\displaystyle DA} (i.e. we get ${\displaystyle \det(DA-BC)}${\displaystyle \det(DA-BC)}) and if ${\displaystyle AB=BA}${\displaystyle AB=BA}, then we should have ${\displaystyle \det(DA-CB)}${\displaystyle \det(DA-CB)}. Note for the last two results, you have to use commutativity of the underlying ring, but not for the first two.

The Surgeon of Death (talk) 14:36, 12 May 2025 (UTC) [reply ]

I was definitely relying entirely on the sources cited for each of the properties mentioned. Since that sentence did not have a source cited inline next to it, I don't remember why I put that in, if I did, although it may be something I got confused about while reading the sources cited for the surrounding sentences. Thiagovscoelho (talk) 13:09, 9 June 2025 (UTC) [reply ]

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