Solve System of Equations in R
Last Updated :
23 Aug, 2021
In this article, we will discuss how to solve a system of equations in R Programming Language.
solve() function in R Language is used to solve the equation. Here equation is like a*x = b, where b is a vector or matrix and x is a variable whose value is going to be calculated.
Syntax: solve(a, b)
Parameters:
- a: coefficients of the equation
- b: vector or matrix of the equation
Example 1: Solving system equation of three equations
Given Equations:
x + 2y + 3z = 20
2x + 2y + 3z = 100
3x + 2y + 8z = 200
Matrix A and B for solution using coefficient of equations:
A->
1 2 3
2 2 3
3 2 8
B->
20
100
200
To solve this using two matrices in R we use the following code:
R
A <- rbind ( c (1, 2, 3),
c (2, 2, 3),
c (3, 2, 8))
B <- c (20, 100, 200)
solve (A, B)
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Output:
80 -36 3.99999999999999
which means x=80, y=-36 and z=4 is the solution for linear equations.
Example 2: Solving system equation of three equations
To get solutions in form of fractions, we use library MASS in R Language and wrap solve function in fractions.
Given Equations:
19x + 32y + 31z = 1110
22x + 28y + 13z = 1406
31x + 12y + 81z = 3040
Matrix A and B for solution using coefficient of equations:
A->
19 32 31
22 28 13
31 12 81
B->
1110
1406
3040
To solve this using two matrices in R we use the following code:
R
library (MASS)
A <- rbind ( c (19, 32, 31),
c (22, 28, 31),
c (31, 12, 81))
B <- c (1110, 1406, 3040)
fractions ( solve (A, B))
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Output:
[1] 159950/2243 -92039/4486 29784/2243
which means x=159950/2243 , y=-92039/4486 and z=29784/2243 is the solution for the above given linear equation.
Example 3: Solving Inverse matrix
R
A <- matrix ( c (4, 7, 3, 6), ncol = 2)
print (A)
print ( "Inverse matrix" )
print ( solve (A))
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Output:
[,1] [,2]
[1,] 4 3
[2,] 7 6
[1] "Inverse matrix"
[,1] [,2]
[1,] 2.000000 -1.000000
[2,] -2.333333 1.333333
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