求z=3x1-x2-x3的最大值其中x1-2x2+x3<=11，-4x1+x2+2x3>=3，-2x2+x3=1，同时x1、x2、x3都大于等于零

这是一个线性规划问题，我们可以使用单纯性法求解。

将约束条件和目标函数写成矩阵形式：

max z = [3 -1 -1] [x1]
                    [x2]
                    [x3]

subject to:

    [ 1  -2  1] [x1] <= 11
    [-4   1  2] [x2] >= 3
    [ 0  -2  1] [x3]  = 1

    x1 ≥ 0, x2 ≥ 0, x3 ≥ 0

将不等式约束转化为等式约束：

    x1 - 2x2 + x3 + x4 = 11
   -4x1 +  x2 + 2x3 - x5 = -3
            -2x2 + x3 + x6 = 1

将目标函数和约束条件写成增广矩阵形式：

[ 3 -1 -1  0  0  0  0] [x1]   [ 0]
[ 1 -2  1  1  0  0  0] [x2]   [11]
[-4  1  2  0 -1  0  0] [x3] = [-3]
[ 0  0  1  0  0  1  0] [x4]   [ 0]
[ 0  0  2  0  0  0 -1] [x5]   [ 3]
[ 0 -2  1  0  0  0  1] [x6]   [ 1]

使用单纯性法解出最优解：

[ 3 -1 -1  0  0  0  0] [x1]   [ 0]
[ 1 -2  1  1  0  0  0] [x2]   [11]
[-4  1  2  0 -1  0  0] [x3] = [-3]
[ 0  0  1  0  0  1  0] [x4]   [ 0]
[ 0  0  2  0  0  0 -1] [x5]   [ 3]
[ 0 -2  1  0  0  0  1] [x6]   [ 1]

pivot at (2, 2)

[ 3 -1 -1  0  0  0  0] [x1]   [     7]
[ 0 -3  2  1  0  0  0] [x2]   [    18]
[ 0 -3  6  0 -1  0  0] [x3] = [   -33]
[ 0  0  1  0  0  1  0] [x4]   [     0]
[ 0  0  0  0  0 -2  1] [x5]   [     9]
[ 0  0  3  0  0  0  1] [x6]   [    12]

pivot at (2, 3)

[ 3 -1 -1  0  0  0  0] [x1]   [ 4.5]
[ 0 -3  2  1  0  0  0] [x2]   [ 6.5]
[ 0  0  4  3 -1  0  0] [x3] = [-4.5]
[ 0  0  1  0  0  1  0] [x4]   [   0]
[ 0  0  0  0  0 -2  1] [x5]   [   9]
[ 0  0  3  0  0  0  1] [x6]   [  12]

pivot at (3, 4)

[ 3 -1 -1  0  0  0  0] [x1]   [ 4.5]
[ 0 -3  0  1  1  0  0] [x2]   [ 7.5]
[ 0  0  4  3 -1  0  0] [x3] = [-4.5]
[ 0  0  1  0  0  1  0] [x4]   [   0]
[ 0  0  0  0  0 -2  1] [x5]   [   9]
[ 0  0  0 -3  3  0  1] [x6]   [  12]

pivot at (2, 5)

[ 3 -1 -1  0  0  0  0] [x1]   [   7]
[ 0 -3  0  1  1  0  0] [x2]   [ 7.5]
[ 0  1  4  3 -3  0  0] [x3] = [ 4.5]
[ 0  0  1  0  0  1  0] [x4]   [   0]
[ 0  0  0  0 -2 -2  1] [x5]   [  -9]
[ 0  0  0 -3  3  0  1] [x6]   [  12]

pivot at (1, 3)

[ 3  0 -2  1  1  0  0] [x1]   [   9]
[ 0 -3  0  1  1  0  0] [x2]   [ 7.5]
[ 0  1  4  3 -3  0  0] [x3] = [ 4.5]
[ 0  0  1  0  0  1  0] [x4]   [   0]
[ 0  0  0  0 -2 -2  1] [x5]   [  -9]
[ 0  0  0 -3  3  0  1] [x6]   [  12]

pivot at (3, 2)

[ 3  0  0  4 -2  0  0] [x1]   [  18]
[ 0 -3  0  1  1  0  0] [x2]   [ 7.5]
[ 0  1  1 -1  1  0  0] [x3] = [ 1.5]
[ 0  0  1  0  0  1  0] [x4]   [   0]
[ 0  0  0  0 -2 -2  1] [x5]   [  -9]
[ 0  0  0 -3  3  0  1] [x6]   [  12]

pivot at (3, 3)

[ 3  0  0  4 -2  0  0] [x1]   [  18]
[ 0 -3  0  1  1  0  0] [x2]   [ 7.5]
[ 0  1  1 -1  1  0  0] [x3] = [ 1.5]
[ 0  0  1  0  0  1  0] [x4]   [   0]
[ 0  0  0  0 -2 -2  1] [x5]   [  -9]
[ 0  0  0 -3  3  0  1] [x6]   [  12]

最优解为：x1=18, x2=0, x3=1.5，最大值为z=3x1-x2-x3=54.5。