由对赌博的思考引起的关于线性不等式组的问题。

对于一个线性不等式组
AX > B　　　　　　　　　　　　　　　　　　　　　　（１）
其中X = [x1 x2 ... xn]T， A(n*n)，B(1*n)为确定的实矩阵。

1。任意给定向量 K = [k1 k2 ... kn]，如何确定MX的符号
2。对于向量 L = [l1(t) l2(t) ... ln(t)]T，其中ln(t)=ln1 x + ln2 (ln1, ln2是确定的实数），如何确定满足（１）的t的范围。
3。一般情况，对于 M = [M1(t) M2(t) ... Mn(t)]，其中 Mn(t)是关于的确定代数函数，如何确定满足（１）的t的范围。

哪位知道要解决这三个问题应该阅读哪些方面的书籍？谢谢！

1。任意给定向量 K = [k1 k2 ... kn]，如何确定MX的符号

how is K related to MX

2。对于向量 L = [l1(t) l2(t) ... ln(t)]T，其中ln(t)=ln1 x + ln2 (ln1, ln2是确定的实数），如何确定满足（１）的t的范围。

how is L related to equation (1), and what is T here?

3。一般情况，对于 M = [M1(t) M2(t) ... Mn(t)]，其中 Mn(t)是关于的确定代数函数，如何确定满足（１）的t的范围。

please explain "Mn(t)是关于的确定代数函数"

1. MX Should be KX

2. Substitute X with L, we get a group of linear inequations for t only. (Should be ln(t) = ln1 t + ln2). t is an real variable.

3. Mn(t) will be some function regarding to t but no longer polynomial or linear, like exp(t) + 2sin(t), or t^5 + ln(t)....

your number 3 has this function M, but your equation does not contain M, what are you talking about?

What will be the range of t which makes AX > B

is your matrix A semi-definite positive or semi-definite negative?

does A has that kind of property? if it does, it is still sovable...

AX > B
is not equivalent to
X > A^-1 * B
even if A is PD.

Since a > b , c > d => a + c > b + d but not vice versa.

Note what I said, for equation: if a = b then c = d <=> a+b=c+d
but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d

A simple counterexample:
A = [ 1 1 ; 1 2] is PD B = [2;3] and
AX > B (1)
But we can't say
X > A^-1 * b = [1;1] (2)

Since obviously x1 = 0, x2 = 3 fits (1) but not (2)

you define AX>B as every element of A is greater than every element of B?

I don't think that's a good definition....

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.

that's solving simultaneous inequalities, then why do u need to generate to matrix form?

use geometric solutions

Geometrical method is useful but limited to 3 variables (and for 3 variables it is not so obvious to see it). Thus some analytical or at least numerical solution is needed.

As long as Mn(t) is in the form as show above, it can be treated as linear. The last equation looks hard to solve, but, i guess it is already another problem.