Timeline for Should the formula for the inverse of a 2x2 matrix be obvious?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 7 at 8:16 | comment | added | Geoffrey Irving | This answer is wonderful! Though as a nit: it seems to use both juxtaposition and $\cdot$ as notation for scalar $\times$ matrix multiplication. | |
| Feb 26, 2012 at 5:55 | comment | added | Noam D. Elkies | @F.Thorne: Thank you! And I see that I should also thank you for not accepting my answer, which made it eligible for a gold star... | |
| Feb 26, 2012 at 5:53 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Made the invocation of Cayley-Hamilton explicit, and linked it to Wikipedia.
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| Feb 26, 2012 at 5:49 | comment | added | Noam D. Elkies | [cont'd] ..., asking not to repeat old standards like logarithm/algorithm, $\int/\Delta$, and the Banach-Tarski joke. Make it community wiki, and hope some good examples get posted before the question gets closed. | |
| Feb 26, 2012 at 5:49 | comment | added | Noam D. Elkies | @Elizabeth S. Q. Goodman: thanks! :-) Linear-algebra anagrams, though? My heuristic for finding "list anagrams" via lattice basis reduction is linear algebra of a kind, but that's surely not what you meant. The closest I can come is something like "label ${\bf R} \oplus {\bf R}$ again", which is what a ${\rm GL}_2({\bf R})$ matrix does, and is an anagram of "linear alg$\oplus$bra". Likewise "label ${\bf R}^e/{\bf R}$ again", which works exactly if I may ignore the "/". Otherwise, try posting a "What are some good math anagrams?" question to mathoverflow ... | |
| Feb 21, 2012 at 15:12 | comment | added | Frank Thorne | That is awesome. | |
| Feb 21, 2012 at 7:37 | comment | added | Elizabeth S. Q. Goodman | Noam, you win Linear Algebra. (To supplement this, maybe you could provide an entertaining linear-algebra-related anagram or two.) | |
| Feb 21, 2012 at 6:55 | comment | added | Guillermo Mantilla | I sometimes give this and the $3 \times 3$ analog of this formula as an exercise; If A is an invertible $3 \times 3$ matrix then $A^{-1}=\Delta^{-1}(A^2-tA +\frac{t^2-s}{2}I)$ where $s=tr(A^2)$, and secretly I'm assuming $1 \neq =-1$. | |
| Feb 21, 2012 at 4:38 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |