Solving simultaneous equations with R

Question

Suppose I have the following equations:

 x + 2y + 3z = 20  
2x + 5y + 9z = 100  
5x + 7y + 8z = 200

How do I solve these equations for x, y and z? I would like to solve these equations, if possible, using R or any other computer tools.

also, I think "ternary" may not be the most descriptive term. I would call this "a set of three coupled linear equations" — Ben Bolker, Nov 16 '11 at 02:15

MYaseen208 · Answer 1 · 2011-11-16T15:45:18.973

33

This should work

A <- matrix(data=c(1, 2, 3, 2, 5, 9, 5, 7, 8), nrow=3, ncol=3, byrow=TRUE)    
b <- matrix(data=c(20, 100, 200), nrow=3, ncol=1, byrow=FALSE)
round(solve(A, b), 3)

     [,1]
[1,]  320
[2,] -360
[3,]  140

edited Nov 16 '11 at 15:45

answered Nov 16 '11 at 06:32

MYaseen208

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If you plug the values 120, 0, -20 back into the equations this is incorrect. It is correct if `byrow = TRUE`. – John Nov 16 '11 at 15:07
Did you? It does still have `byrow = FALSE` – Marcin Jan 25 '16 at 15:42
@MYaseen208 What if my system of equations is transcendental? – Plinth Oct 14 '16 at 16:59
Is solve the most efficient way to do this? What if the matrix is very big? Or the condition is low? – wolfsatthedoor Apr 28 '18 at 17:52

score 13 · Answer 2 · answered Dec 25 '14 at 19:28

For clarity, I modified the way the matrices were constructed in the previous answer.

a <- rbind(c(1, 2, 3), 
           c(2, 5, 9), 
           c(5, 7, 8))
b <- c(20, 100, 200)
solve(a, b)

In case we need to display fractions:

library(MASS)
fractions(solve(a, b))

banbh · Answer 3 · 2018-03-02T14:26:29.590

5

Another approach is to model the equations using lm as follows:

lm(b ~ . + 0, 
   data = data.frame(x = c(1, 2, 5), 
                     y = c(2, 5, 7), 
                     z = c(3, 9, 8), 
                     b = c(20, 100, 200)))

which produces

Coefficients:
   x     y     z  
 320  -360   140

If you use the tibble package you can even make it read just like the original equations:

lm(b ~ . + 0, 
   tibble::tribble(
     ~x, ~y, ~z,  ~b,
      1,  2,  3,  20,
      2,  5,  9, 100,
      5,  7,  8, 200))

which produces the same output.

edited Mar 02 '18 at 14:26

answered Feb 22 '18 at 21:15

banbh

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Is there any benefit to this method? Precision? Computation speed? – Gimelist Apr 27 '20 at 14:11
I have not done any benchmarking, but if I were to guess directly using matrices is probably the fastest. However, I use the `lm` approach if it helps explain the purpose of my code. This is clearly a vague criterion; in my case if `x`, `y`, and `z` are variables or factors in my analysis then I may use `lm`. – banbh Apr 28 '20 at 14:52

score -3 · Answer 4 · edited Mar 15 '17 at 11:17

-3

A <- matrix(data=c(1, 2, 3, 2, 5, 9, 5, 7, 8),nrow=3,ncol=3,byrow=TRUE)    
b <- matrix(data=c(20, 100, 200),nrow=3,ncol=1,byrow=FALSE)
solve(A)%*% b

Note that this is a square matrix!

edited Mar 15 '17 at 11:17

Sampada

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answered Mar 15 '17 at 11:04

Josephine

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how is this substantively different from the previously posted answers? (note that `solve(A,b)` is equivalent to but more efficient than `solve(A) %*% b`) – Ben Bolker Mar 15 '17 at 11:26

Solving simultaneous equations with R

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