library(matlib)EXAMPLE 1: Then we create the coefficient matrix and the vector for the right hand side of the system of linear equations.
A <- matrix(c(0, -1, 1, 0, 1, 1, 0, 1, 3, -4, 2, 0, -1, 0, 4, -4), 4, 4)
b <- c(1, 1, 5, -2)Here, we did not type “nrow=” and “ncol=”. If it is clear we can skip typying them. Using the showEqn() function we can see the system of linear equations.
showEqn(A, b)## 0*x1 + 1*x2 + 3*x3 - 1*x4 = 1
## -1*x1 + 1*x2 - 4*x3 + 0*x4 = 1
## 1*x1 + 0*x2 + 2*x3 + 4*x4 = 5
## 0*x1 + 1*x2 + 0*x3 - 4*x4 = -2Now using the echelon() function we can see how Gaussian Elimination works. With this example, we can type in R as:
echelon(A, b, verbose=TRUE, fractions=TRUE)##
## Initial matrix:
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 1 3 -1 1
## [2,] -1 1 -4 0 1
## [3,] 1 0 2 4 5
## [4,] 0 1 0 -4 -2
##
## row: 1
##
## exchange rows 1 and 2
## [,1] [,2] [,3] [,4] [,5]
## [1,] -1 1 -4 0 1
## [2,] 0 1 3 -1 1
## [3,] 1 0 2 4 5
## [4,] 0 1 0 -4 -2
##
## multiply row 1 by -1
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -1 4 0 -1
## [2,] 0 1 3 -1 1
## [3,] 1 0 2 4 5
## [4,] 0 1 0 -4 -2
##
## subtract row 1 from row 3
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -1 4 0 -1
## [2,] 0 1 3 -1 1
## [3,] 0 1 -2 4 6
## [4,] 0 1 0 -4 -2
##
## row: 2
##
## multiply row 2 by 1 and add to row 1
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 7 -1 0
## [2,] 0 1 3 -1 1
## [3,] 0 1 -2 4 6
## [4,] 0 1 0 -4 -2
##
## subtract row 2 from row 3
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 7 -1 0
## [2,] 0 1 3 -1 1
## [3,] 0 0 -5 5 5
## [4,] 0 1 0 -4 -2
##
## subtract row 2 from row 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 7 -1 0
## [2,] 0 1 3 -1 1
## [3,] 0 0 -5 5 5
## [4,] 0 0 -3 -3 -3
##
## row: 3
##
## multiply row 3 by -1/5
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 7 -1 0
## [2,] 0 1 3 -1 1
## [3,] 0 0 1 -1 -1
## [4,] 0 0 -3 -3 -3
##
## multiply row 3 by 7 and subtract from row 1
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 6 7
## [2,] 0 1 3 -1 1
## [3,] 0 0 1 -1 -1
## [4,] 0 0 -3 -3 -3
##
## multiply row 3 by 3 and subtract from row 2
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 6 7
## [2,] 0 1 0 2 4
## [3,] 0 0 1 -1 -1
## [4,] 0 0 -3 -3 -3
##
## multiply row 3 by 3 and add to row 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 6 7
## [2,] 0 1 0 2 4
## [3,] 0 0 1 -1 -1
## [4,] 0 0 0 -6 -6
##
## row: 4
##
## multiply row 4 by -1/6
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 6 7
## [2,] 0 1 0 2 4
## [3,] 0 0 1 -1 -1
## [4,] 0 0 0 1 1
##
## multiply row 4 by 6 and subtract from row 1
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 1
## [2,] 0 1 0 2 4
## [3,] 0 0 1 -1 -1
## [4,] 0 0 0 1 1
##
## multiply row 4 by 2 and subtract from row 2
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 1
## [2,] 0 1 0 0 2
## [3,] 0 0 1 -1 -1
## [4,] 0 0 0 1 1
##
## multiply row 4 by 1 and add to row 3
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 1
## [2,] 0 1 0 0 2
## [3,] 0 0 1 0 0
## [4,] 0 0 0 1 1Therefore the solution for this system is x1 = 1, x2 = 2, x3 = 0, x4 = 1
EXAMPLE 2:
A <- matrix(c(0, -9, 4, 4, 5, 6, 3, 8, 4, 6, 0, 4, 7, 7, 5, 12), 4, 4)
b <- c(-9, -17, -3, -12)Using the showEqn() function we can see the system of linear equations.
showEqn(A, b)## 0*x1 + 5*x2 + 4*x3 + 7*x4 = -9
## -9*x1 + 6*x2 + 6*x3 + 7*x4 = -17
## 4*x1 + 3*x2 + 0*x3 + 5*x4 = -3
## 4*x1 + 8*x2 + 4*x3 + 12*x4 = -12Now using the echelon() function we obtain the reduced echelon form of the augmented matrix. From this we have the reduced echelon form of the augmented matrix as follows:
echelon(A, b, verbose=TRUE, fractions=TRUE)##
## Initial matrix:
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 5 4 7 -9
## [2,] -9 6 6 7 -17
## [3,] 4 3 0 5 -3
## [4,] 4 8 4 12 -12
##
## row: 1
##
## exchange rows 1 and 2
## [,1] [,2] [,3] [,4] [,5]
## [1,] -9 6 6 7 -17
## [2,] 0 5 4 7 -9
## [3,] 4 3 0 5 -3
## [4,] 4 8 4 12 -12
##
## multiply row 1 by -1/9
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -2/3 -2/3 -7/9 17/9
## [2,] 0 5 4 7 -9
## [3,] 4 3 0 5 -3
## [4,] 4 8 4 12 -12
##
## multiply row 1 by 4 and subtract from row 3
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -2/3 -2/3 -7/9 17/9
## [2,] 0 5 4 7 -9
## [3,] 0 17/3 8/3 73/9 -95/9
## [4,] 4 8 4 12 -12
##
## multiply row 1 by 4 and subtract from row 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -2/3 -2/3 -7/9 17/9
## [2,] 0 5 4 7 -9
## [3,] 0 17/3 8/3 73/9 -95/9
## [4,] 0 32/3 20/3 136/9 -176/9
##
## row: 2
##
## exchange rows 2 and 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -2/3 -2/3 -7/9 17/9
## [2,] 0 32/3 20/3 136/9 -176/9
## [3,] 0 17/3 8/3 73/9 -95/9
## [4,] 0 5 4 7 -9
##
## multiply row 2 by 3/32
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 -2/3 -2/3 -7/9 17/9
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 17/3 8/3 73/9 -95/9
## [4,] 0 5 4 7 -9
##
## multiply row 2 by 2/3 and add to row 1
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1/4 1/6 2/3
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 17/3 8/3 73/9 -95/9
## [4,] 0 5 4 7 -9
##
## multiply row 2 by 17/3 and subtract from row 3
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1/4 1/6 2/3
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 0 -7/8 1/12 -1/6
## [4,] 0 5 4 7 -9
##
## multiply row 2 by 5 and subtract from row 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1/4 1/6 2/3
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 0 -7/8 1/12 -1/6
## [4,] 0 0 7/8 -1/12 1/6
##
## row: 3
##
## exchange rows 3 and 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1/4 1/6 2/3
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 0 7/8 -1/12 1/6
## [4,] 0 0 -7/8 1/12 -1/6
##
## multiply row 3 by 8/7
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1/4 1/6 2/3
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 0 1 -2/21 4/21
## [4,] 0 0 -7/8 1/12 -1/6
##
## multiply row 3 by 1/4 and add to row 1
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 1/7 5/7
## [2,] 0 1 5/8 17/12 -11/6
## [3,] 0 0 1 -2/21 4/21
## [4,] 0 0 -7/8 1/12 -1/6
##
## multiply row 3 by 5/8 and subtract from row 2
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 1/7 5/7
## [2,] 0 1 0 31/21 -41/21
## [3,] 0 0 1 -2/21 4/21
## [4,] 0 0 -7/8 1/12 -1/6
##
## multiply row 3 by 7/8 and add to row 4
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 1/7 5/7
## [2,] 0 1 0 31/21 -41/21
## [3,] 0 0 1 -2/21 4/21
## [4,] 0 0 0 0 0
##
## row: 4A <- matrix(c(1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0,
0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0,
0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0,
1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1,
0, 0, 0, 1 ), nrow = 8, ncol = 16, byrow = TRUE)which is a 8 × 16 matrix and the vector for the right hand side of the system of linear equations is
b <- c(21.5, 20.9, 33.4, 22.1, 4.5, 14.2, 42.4, 62.9)Then we use the Solve() function. The following is the output from R:
Solve(A,b)## x1 - 1*x6 - 1*x7 - 1*x8 - 1*x10 - 1*x11 - 1*x12 - 1*x14 - 1*x15 - 1*x16 = -71.9
## x2 + x6 + x10 + x14 = 14.2
## x3 + x7 + x11 + x15 = 42.4
## x4 + x8 + x12 + x16 = 36.8
## x5 + x6 + x7 + x8 = 20.9
## x9 + x10 + x11 + x12 = 33.4
## x13 + x14 + x15 + x16 = 22.1
## 0 = 26.1When we look at the last equation, it says 0 = 26.1, which does not make sense. This is the form in Theorem 1.6. Therefore, there is no solution to the system of linear equations. Thus, we conclude that there does not exist such a table. This means that this perturbed information does not make sense for a statistical analysis for studying relationship. Thus perturbation does not work for releasing this information.