R 行简化阶梯型

R中是否有函数生成矩阵的

缩减行梯队形式

？。参考资料说没有。你同意吗？

看起来没有内置的功能，但我在第页找到了这个rref函数

---

rrefpracma包还包含一个实现。请参见pracma:：rref。
我没有足够的代表发表评论，但上面由soldier.moth在接受答案[edit 2018:不再是接受答案]中给出的函数有缺陷-它不处理rref解决方案主对角线上有零的矩阵。试试看

m最近还开发了一个用于教学线性代数（）的软件包，该软件包既计算矩阵的梯队形式，又显示了沿途使用的步骤

示例来自：

库（'matlib'）
A这并不能回答这个问题。若要评论或要求作者澄清，请在他们的帖子下方留下评论-你可以随时在自己的帖子上发表评论，一旦你有足够的评论，你就可以发表评论了。对不起，我是新来的，可能遗漏了一些东西，但是：士兵提供的“公认答案”。上面的moth是buggy（当我自己尝试使用它时，我发现了困难的方法！）所以我认为这一点很重要。我没有足够的代表对士兵发表评论。莫思的答案是直接的，所以我创造了一个新的答案-我应该在这里做什么？获得足够的声誉首先发表评论？耸耸肩我真的不认为有人会因为我指出一个四年没有纠正的不明显的错误而烦恼n、 考虑到上面发布的错误“解决方案”被接受为答案，我很困惑为什么同一代码的固定版本被认为不那么重要。@GeoffreyBrent，欢迎来到Stackoverflow。在这里，规则取代了常识。
 rref <- function(A, tol=sqrt(.Machine$double.eps),verbose=FALSE,
                 fractions=FALSE){
  ## A: coefficient matrix
  ## tol: tolerance for checking for 0 pivot
  ## verbose: if TRUE, print intermediate steps
  ## fractions: try to express nonintegers as rational numbers
  ## Written by John Fox
  if (fractions) {
    mass <- require(MASS)
    if (!mass) stop("fractions=TRUE needs MASS package")
  }
  if ((!is.matrix(A)) || (!is.numeric(A)))
    stop("argument must be a numeric matrix")
  n <- nrow(A)
  m <- ncol(A)
  for (i in 1:min(c(m, n))){
    col <- A[,i]
    col[1:n < i] <- 0
    # find maximum pivot in current column at or below current row
    which <- which.max(abs(col))
    pivot <- A[which, i]
    if (abs(pivot) <= tol) next     # check for 0 pivot
    if (which > i) A[c(i, which),] <- A[c(which, i),]  # exchange rows
    A[i,] <- A[i,]/pivot            # pivot
    row <- A[i,]
    A <- A - outer(A[,i], row)      # sweep
    A[i,] <- row                    # restore current row
    if (verbose)
      if (fractions) print(fractions(A))
      else print(round(A,round(abs(log(tol,10)))))
  }
  for (i in 1:n)
    if (max(abs(A[i,1:m])) <= tol)
      A[c(i,n),] <- A[c(n,i),] # 0 rows to bottom
  if (fractions) fractions (A)
  else round(A, round(abs(log(tol,10))))
}

rref <- function(A, tol=sqrt(.Machine$double.eps),verbose=FALSE,
                 fractions=FALSE){
  ## A: coefficient matrix
  ## tol: tolerance for checking for 0 pivot
  ## verbose: if TRUE, print intermediate steps
  ## fractions: try to express nonintegers as rational numbers
  ## Written by John Fox
  # Modified by Geoffrey Brent 2014-12-17 to fix a bug
  if (fractions) {
    mass <- require(MASS)
    if (!mass) stop("fractions=TRUE needs MASS package")
  }
  if ((!is.matrix(A)) || (!is.numeric(A)))
    stop("argument must be a numeric matrix")
  n <- nrow(A)
  m <- ncol(A)
  x.position<-1
  y.position<-1
  # change loop:
  while((x.position<=m) & (y.position<=n)){
    col <- A[,x.position]
    col[1:n < y.position] <- 0
    # find maximum pivot in current column at or below current row
    which <- which.max(abs(col))
    pivot <- col[which]
    if (abs(pivot) <= tol) x.position<-x.position+1     # check for 0 pivot
    else{
      if (which > y.position) { A[c(y.position,which),]<-A[c(which,y.position),] } # exchange rows
      A[y.position,]<-A[y.position,]/pivot # pivot
      row <-A[y.position,]
      A <- A - outer(A[,x.position],row) # sweep
      A[y.position,]<-row # restore current row
      if (verbose)
        if (fractions) print(fractions(A))
        else print(round(A,round(abs(log(tol,10)))))
      x.position<-x.position+1
      y.position<-y.position+1
    }
  }
  for (i in 1:n)
    if (max(abs(A[i,1:m])) <= tol)
      A[c(i,n),] <- A[c(n,i),] # 0 rows to bottom
  if (fractions) fractions (A)
  else round(A, round(abs(log(tol,10))))
}

library('matlib')
A <- matrix(c(2, 1, -1,-3, -1, 2,-2, 1, 2), 3, 3, byrow=TRUE)
b <- c(8, -11, -3)
echelon(A, b, verbose=TRUE, fractions=TRUE)

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