Timeline for Calculate the inverse of a matrix
Current License: CC BY-SA 4.0
12 events
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| Feb 24, 2021 at 23:08 | history | bounty ended | Bubbler | ||
| Feb 19, 2021 at 20:11 | comment | added | William Martens | THIS IS EFFORT. Really; this is something I love - really good work! :o | |
| Oct 22, 2020 at 21:47 | comment | added | Sisyphus |
@TobiasKnauss This was an answer intended to display a neat method without builtins. Of course you could do inv, which is much shorter, but also (imo) much more boring :)
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| Oct 22, 2020 at 17:38 | comment | added | Tobias Knauss |
Why not simply use inv(M)???
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| Oct 21, 2020 at 23:21 | history | edited | Sisyphus | CC BY-SA 4.0 |
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| Oct 21, 2020 at 14:23 | comment | added | Giuseppe | This is sweet. Wikipedia's page for Matrix inverses notes this formula as a version of Newton's Method but its references are pretty useless for finding \$V_0\$. | |
| Oct 21, 2020 at 5:06 | history | edited | Sisyphus | CC BY-SA 4.0 |
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| Oct 21, 2020 at 4:45 | comment | added | Sisyphus | Thanks @xnor, that's a significant improvement over explicitly constructing the identity matrix. | |
| Oct 21, 2020 at 4:43 | history | edited | Sisyphus | CC BY-SA 4.0 |
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| Oct 21, 2020 at 3:16 | comment | added | xnor |
Thanks for sharing this! I didn't know you there was an simple iterative scheme that's guaranteed to converge for an easy-to-compute start value. I know this answer isn't here for golfing, but it looks like you can write V=2*V-V*A*V. There's always writing A^0 for the identity in V*=2*A^0-A*V, but that's one longer.
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| Oct 21, 2020 at 2:03 | history | edited | Sisyphus | CC BY-SA 4.0 |
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| Oct 21, 2020 at 1:57 | history | answered | Sisyphus | CC BY-SA 4.0 |