12:04 AM - 6LACK Tripbwai (1): im gay
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SteamID64 | 76561198109904512 |
SteamID3 | [U:1:149638784] |
SteamID32 | STEAM_0:0:74819392 |
Country | Netherlands |
Signed Up | October 7, 2013 |
Last Posted | September 20, 2025 at 11:39 PM |
Posts | 2810 (0.6 per day) |
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wait a second once you get XA(A + I) = A + I isnt that just the solution
doesnt matter if A+I is invertible or not, you know XA=I because I is the only matrix that would solve IC=C for C=(A+I) unless C=0
so since XA=I you can multiply both sides by A^-1 and get X=A^-1
i made this way too complicated in my last post rofl
of course theres still a case where A+I=0 where X can equal any matrix it wants
edit im retarded
ok after thinking some more. if A is invertible, then it is row equivalent to I. you would think adding two row equivalent matrices would be row equivalent so A would be row equivalent to (A+I), but theres actually an exception here...when A has a row which is equal to the same row in -I, then adding that to I will mean that to get to (A+I) from A, you must perform the row operation of adding the negative of a row to itself, which is the same as multiplying by 0. this isnt a valid row operation so they aren't row equivalent. however this is the only exception.
so this means that either (A+I) is invertible, or it has at least one row of all 0's.
plugging this back into the equation XA(A + I) = A + I doesn't really help though. if all of the rows are 0 (A=-I), then it obviously solves for all X, but if only some rows are 0, I can't think of anything that would mean, especially since A+I is on the right side of the multiplication. if it was on the left side then that would mean an entire row of the product would also be 0 which would be equal to A+I, but on the right side that doesnt seem to necessarily be the case. you can use distributive property on other side to prove (A+I)A has the same zero rows, but you’re still left multiplying that by X so it doesnt help.
GrapeJuiceIIINot very related to the variance but i think fall damage should be calculated by time in the air rather than speed if it’s even possible. it would eliminate being cratered by rockets and by people just landing on you as you’re falling.
bad idea you would take twice as much damage from rocket jumping as falling down from a ledge or something
greatest match of LAN dominates the single map records
please remove the kicking for inactivity, theres no point since there isnt a sub system, and it kicks people during pauses
scooby doo night of 100 frights is the best 3d platformer of all time
nicest person in tf2
until you bring up overwatch, her hatred for it is stronger than the fires of hell...
im fucked up too ive never really gotten sick before in my life either
Jasbutts200most importantly, anyone got photos of yight wearing fedora? are those only available for donors?
I have quite a few.
upload your lan pics pls!