CODEWITHSUNDEEP
r
nilsinelabore
nilsinelabore

Reputation: 5095

Solving simultaneous equation in R

I want to solve the following simultaneous equations using R:

# -0.9x+0.01y+0.001z+0.001n=0
# 0.9x-0.82y+0.027z+0.027n=0
# 0x+0.81y+(0.243-1)z+0.243n=0
# 0x+0y+0.729z+(0.729-1)n=0
# x+y+z+n=1

I tried:

A <- matrix(data=c(-0.9,0.01,0.001,0.001,0.9,-0.82,0.027,0.027,0,0.81,0.243-1,0.243,0,0,0.729,0.729-1,1,1,1,1), nrow=5, ncol=4, byrow=TRUE)    
b <- matrix(data=c(0, 0, 0, 0, 1), nrow=5, ncol=1, byrow=FALSE)
round(solve(A, b), 4)

and it caught error:

Error in solve.default(A, b) : 'a' (5 x 4) must be square

What does the error mean? Is this function only applicable to square matrix?

Edit:

I removed the last equation and tried:


A <- matrix(data=c(-0.9,0.01,0.001,0.001,0.9,-0.82,0.027,0.027,0,0.81,0.243-1,0.243,0,0,0.729,0.729-1), nrow=4, ncol=4, byrow=TRUE)    
b <- matrix(data=c(0, 0, 0, 0), nrow=4, ncol=1, byrow=FALSE)
A;b;
round(solve(A, b), 4)

which caught error:

Error in solve.default(A, b) : Lapack routine dgesv: system is exactly singular: U[4,4] = 0

Upvotes: 0

Views: 209

Answers (1)

Ben Bolker
Ben Bolker

Reputation: 226172

Since you have 5 linear equations (rows of your matrix) and only 4 columns (variables), then your problem is overdetermined and in general it can't be solved. Consider a smaller example: x+y=1; x-2*y=2; 3*x-5=0. Once you use the first two equations:

(1) x+y=1        → y=1-x
(2) x-2*(1-x)=2  → 3*x-2=2 → x=4/3  ## substitute (1) into second eq.
(3) → y=-1/3

you have a solution for x and y (x=4/3, y=-1/3), so the last equation is either redundant or makes your solution impossible (in this case, the latter: 3*x-5 = -1 != 5).

On the other hand, if you have fewer rows than columns, then your system is underdetermined and you can't get a unique answer.

So the number of matrix rows must equal the number of columns → you can only do this with a square matrix. This is not a limitation of the R function, it's about the underlying math of linear equations ...

In your specific case the problem is that the sum of the first four rows of your matrix add up to zero, so these are redundant. As you found out, just dropping the last equation doesn't help. But if you drop one of the redundant rows you can get an answer.

solve(A[-4,],b[-4])

Note you get the same answer whether you drop row 1, 2, 3, or 4.

Upvotes: 2

Related Questions