# [R] Solutions of equation systems

Berend Hasselman bhh at xs4all.nl
Sat Aug 15 16:01:46 CEST 2009

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Moreno Mancosu wrote:
>
> Dear Greg,
>
> I tried to use function "solve", but I have some problems with it.
> Let's try with a simple equation:
>
> x + 3y -2z = 5
> 3x + 5y + 6z =7
>
>  > a<-matrix(c(1,3,-2,3,5,6),nrow=2,ncol=3)
>  > a
>      [,1] [,2] [,3]
> [1,]    1   -2    5
> [2,]    3    3    6
>  > b<-matrix(c(5,7),nrow=2,ncol=1)
>  > b
>      [,1]
> [1,]    5
> [2,]    7
>
> When I try to solve this system, I find this error:
>
>  > solve(a,b)
> Error in solve.default(a, b) : 'b' must be compatible with 'a'
>
> Also, when i try to solve a system with n equation and n variables,
> solve works perfectly.
>

If you do ?solve then you will see that
the matrix a must be non square (it says so in the description of a in
section Arguments)/
and in section Details right at the end it is written that "qr.solve can
handle non-square systems".

Your example can be solved in several ways.
See this

<example>
A <- matrix(c(1,2,3,5,6,7),nrow=2,byrow=T)
b <- c(2,8)

A <- matrix(c(1,3,-2,3,5,6),nrow=2,ncol=3)
b <- matrix(c(5,7),nrow=2,ncol=1)

# a particular solution
xsolve <- qr.solve(A,b)
xsolve

# another particular solution

xany <- solve(qr(A,LAPACK=T),b)
xany

# QR decomposition of A-transposed

ATQR <- qr(t(A),LAPACK=TRUE)
ATQR

qr.Q(ATQR)
qr.R(ATQR)

# minimum norm solution comes from QR decomposition of t(A)
# and solving t(R) %*% t(Q) %*% x = b in two steps
# ! Lapack does a pivoted QR

z <- solve(t(qr.R(ATQR)),b[ATQR\$pivot])
xmin <- qr.Q(ATQR) %*% z
xmin

# null space of A
library(MASS)
A.null <- Null(t(A))
A.null

# test
A %*% (xmin+5*A.null)
</example>

! Use the manual

Berend
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