Angel [EMAIL PROTECTED] writes:
Why do
x-b%*%ginv(A)
and
x-solve(A,b)
give different results?. It seems that I am missing some basic feature of
matrix indexing.
e.g.:
A-matrix(c(0,-4,4,0),nrow=2,ncol=2)
b-c(-16,0)
x-b%*%ginv(A);x
x-solve(A,b);x
[ginv() is from MASS, please remember
Uwe Ligges wrote:
Angel wrote:
Why do
x-b%*%ginv(A)
Why should it be the same???
Here, you are calculating (A+ is G-Inverse)
x = b * A+
Let me add: More exactly, you are calculating
x = b' * A+
because b * A+ doesn't fit and R is somehow intelligent here and
transposes b for you instead
Why do
x-b%*%ginv(A)
and
x-solve(A,b)
give different results?. It seems that I am missing some basic feature of
matrix indexing.
e.g.:
A-matrix(c(0,-4,4,0),nrow=2,ncol=2)
b-c(-16,0)
x-b%*%ginv(A);x
x-solve(A,b);x
Thanks in advance,
Angel
__
[EMAIL
Why do
x-b%*%ginv(A)
and
x-solve(A,b)
give different results?.
they do (in cases the solution to A x = b is unique):
R A - matrix(c(0,-4,4,0),nrow=2,ncol=2)
R A
[,1] [,2]
[1,]04
[2,] -40
R b - c(-16,0)
R x1 - ginv(A) %*% b - NOT b %*% ginv(A)
R x1
[,1]
[1,]
