As I described
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x1 + x2 > 2
x1 + 2x2 > 3

is expressed as AX > B, where A = [ 1 1 ; 1 2 ]; B = [2 3]T

Maybe it is a formal notation but I just would like to solve the problem.
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Yeah!
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请教一个简单的不等式问题 － 香陵居士 　2007-08-31 16:06 (518 bytes , 719reads )
2. － bugzzj 　2007-09-24 18:30 (77 bytes , 377reads )
1. － bugzzj 　2007-09-24 18:09 (76 bytes , 307reads )
please explain your notation － icky 　2007-08-31 17:53 (405 bytes , 302reads )
Sorry... so many typos... － 香陵居士 　2007-08-31 18:18 (276 bytes , 394reads )
I think the examples in 3 can still be treated as linear. － bugzzj 　2007-09-25 10:26 (236 bytes , 366reads )
i'm still confused － icky 　2007-09-06 12:40 (100 bytes , 315reads )
It simply says if each element of X is a function of variable t － 香陵居士 　2007-09-06 15:05 (46 bytes , 255reads )
this is not easy, even for linear function of t － icky 　2007-09-06 17:22 (66 bytes , 245reads )
i mean － icky 　2007-09-06 17:22 (68 bytes , 266reads )
I do understand what you mean but what's tricky here is that － 香陵居士 　2007-09-06 22:34 (120 bytes , 279reads )
if A is positive definite, then Ax > B eq to x > A^-1 B － icky 　2007-09-06 23:49 (0 bytes , 282reads )
... Can you prove that? － 香陵居士 　2007-09-07 02:00 (319 bytes , 262reads )
Thanks, but the most important thing is － 香陵居士 　2007-09-08 09:29 (173 bytes , 256reads )
aiyah.... － icky 　2007-09-08 01:40 (114 bytes , 244reads )
As I described － 香陵居士 　2007-09-08 00:19 (161 bytes , 268reads )
how do u define your AX > B ? AX is a vector, B is also a vector － icky 　2007-09-07 17:23 (117 bytes , 290reads )
oh, ok, you are saying vectors can not be compared directly － icky 　2007-09-07 12:38 (0 bytes , 261reads )