[实际应用] R求解阶梯矩阵的方法 [推广有奖]

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llb_321

发表于 2020-7-16 08:21:50 |AI写论文

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本来想自编个函数，后来觉得R这么强大，包那么多，不应该没有相应的解决办法啊。
找了一堆包，终于找到了，而且非常贴心地能够给出求解的每一步，赞。

小伙伴们，闲言少叙。
关键的包：matlib
重点函数：gaussianElimination()和echelon()

这是官方给的gaussianElimination示例：

A <- matrix(c(2, 1, -1,-3, -1, 2,-2, 1, 2), 3, 3, byrow=TRUE)
b <- c(8, -11, -3)
gaussianElimination(A, b, verbose=TRUE, fractions=TRUE)

执行结果，牛吧：
Initial matrix:
   [,1] [,2] [,3] [,4]
[1,] 2 1 -1 8

[2,]  -3 -1 2  -11

[3,]  -2 1 2 -3

row: 1  exchange rows 1 and 2
   [,1] [,2] [,3] [,4]
[1,]  -3 -1 2  -11

[2,] 2 1 -1 8

[3,]  -2 1 2 -3
multiply row 1 by -1/3
   [,1] [,2] [,3] [,4]
[1,] 1  1/3 -2/3 11/3
[2,] 2 1 -1 8
[3,] -2 1 2 -3

multiply row 1 by 2 and subtract from row 2
   [,1] [,2] [,3] [,4]
[1,] 1  1/3 -2/3 11/3
[2,] 0  1/3  1/3  2/3
[3,] -2 1 2 -3
multiply row 1 by 2 and add to row 3
   [,1] [,2] [,3] [,4]
[1,] 1  1/3 -2/3 11/3
[2,] 0  1/3  1/3  2/3
[3,] 0  5/3  2/3 13/3
row: 2  exchange rows 2 and 3
   [,1] [,2] [,3] [,4]
[1,] 1  1/3 -2/3 11/3
[2,] 0  5/3  2/3 13/3
[3,] 0  1/3  1/3  2/3

multiply row 2 by 3/5
   [,1] [,2] [,3] [,4]
[1,] 1  1/3 -2/3 11/3
[2,] 0 1  2/5 13/5
[3,] 0  1/3  1/3  2/3
multiply row 2 by 1/3 and subtract from row 1
   [,1] [,2] [,3] [,4]
[1,] 1 0 -4/5 14/5
[2,] 0 1  2/5 13/5
[3,] 0  1/3  1/3  2/3

multiply row 2 by 1/3 and subtract from row 3
   [,1] [,2] [,3] [,4]
[1,] 1 0 -4/5 14/5
[2,] 0 1  2/5 13/5
[3,] 0 0  1/5 -1/5
row: 3  multiply row 3 by 5
   [,1] [,2] [,3] [,4]
[1,] 1 0 -4/5 14/5
[2,] 0 1  2/5 13/5
[3,] 0 0 1 -1

multiply row 3 by 4/5 and add to row 1
   [,1] [,2] [,3] [,4]
[1,] 1 0 0 2
[2,] 0 1  2/5 13/5
[3,] 0 0 1 -1

multiply row 3 by 2/5 and subtract from row 2
   [,1] [,2] [,3] [,4]
[1,]  1 0 0 2

[2,]  0 1 0 3

[3,]  0 0 1 -1

用echelon(A,b)直接得到结果。