\[\vec{u} =\begin{bmatrix}0.5 \\ 0.5 \end{bmatrix}\] \[\vec{v} =\begin{bmatrix}3 \\ 4 \end{bmatrix}\] (1) Dot Product
w <- (u[1]*v[1]) + (u[2]*v[2])
w## [1] -0.5length_of_vector_u <-sqrt(sum(u^2))
length_of_vector_u## [1] 0.7071068length_of_vector_v <-sqrt(sum(v^2))
length_of_vector_v## [1] 5linear_combo <- (3*u) - (2*v)
linear_combo ## [1] -4.5 9.5angle <- acos(w/(length_of_vector_u*length_of_vector_v))
angle## [1] 1.712693Set up a system of equations with 3 variables and 3 constraints and solve for x. Please write a function in R that will take two variables (matrix A & constraint vector b) and solve using elimination. Your function should produce the right answer for the system of equations for any 3-variable, 3-equation system….The approach that you should employ is to construct an Upper Triangular Matrix and then back-substitute to get the solution. Alternatively, you can augment the matrix A with vector b and jointly apply the Gauss Jordan elimination procedure.
The function below solves a system of three equations by augmenting matrix A with vector b and jointly applying the Gauss Jordan elimination procedure. The function should return a tuple which provides a solution that can solve all three equations simulataneously.
There are four rules that this function has to follow:
In the code below, I print out the updated matrix after each application of the rules above to ensure that everything is working correctly.
solve_three_by_three_equations <- function(m, v){
#These next two lines augment the matrix with the vector and removes the column names
aug_matrix <- cbind(m,v)
colnames(aug_matrix) <- NULL
print('Below is the original augmented matrix:')
print(aug_matrix)
cat("\n\n")
#Initialize the row and col variables to "1".
col <- 1
row <- 1
#Check if the first coefficient in the first row, first column equals 1. If yes, proceed with the other rules.
if (aug_matrix[row,col] == 1){
#Apply Rules (2) and (3) multiply both sides of an equation by a nonzero constant and replaced by the sum of itself and a multiple of another
if (aug_matrix[row+1,col] != 0){
aug_matrix[row+1,] <- (-aug_matrix[row+1,col] * aug_matrix[row,]) + aug_matrix[row+1,]
#print the updated augmented matrix
print('The first step applied to the augmented matrix:')
print(aug_matrix)
cat("\n\n")
}
#Apply Rules (2) and (3) multiply both sides of an equation by a nonzero constant and replaced by the sum of itself and a multiple of another
if (aug_matrix[row+2,col] !=0){
aug_matrix[row+2,] <- (-aug_matrix[row+2,col] * aug_matrix[row,]) + aug_matrix[row+2,]
#print the updated augmented matrix
print('The first column is now complete:')
print(aug_matrix)
cat("\n\n")
}
#Apply Rule (2) An equation has both sides multiplied by a nonzero constant in order to get a "1" in the second row, second column
if (aug_matrix[row+1,col+1]!=0){
aug_matrix[row+1,] <- (1/aug_matrix[row+1,col+1]) * aug_matrix[row+1,]
#print the updated augmented matrix
print('The second row and column is now complete:')
print(aug_matrix)
cat("\n\n")
}
#Apply Rules (2) and (3) multiply both sides of an equation by a nonzero constant and replaced by the sum of itself and a multiple of another
if (aug_matrix[row+2,col+1] != 0){
aug_matrix[row+2,] <- (-aug_matrix[row+2,col+1] * aug_matrix[row+1,]) + aug_matrix[row+2,]
#print the updated augmented matrix
print('The third row now has a 0 in its second column:')
print(aug_matrix)
cat("\n\n")
}
#For the final step, apply Rule (2) An equation has both sides multiplied by a nonzero constant in order to get a "1" in the third row, third column
if (aug_matrix[row+2,col+2] !=0){
aug_matrix[row+2,] <- (1/aug_matrix[row+2,col+2]) * aug_matrix[row+2,]
#print the final Gauss matrix
print('The final matrix:')
print(aug_matrix)
cat("\n\n")
}
}
#Use backward population to find the values of x1, x2, and x3.
x3 <- aug_matrix[row+2,col+3]
x2 <- aug_matrix[row+1,col+3] - (aug_matrix[row+1,col+2]* aug_matrix[row+2,col+3])
x1 <- aug_matrix[row,col+3] - (aug_matrix[row+1,col+3] - (aug_matrix[row+1,col+2]* aug_matrix[row+2,col+3])) - (3*aug_matrix[row+2,col+3])
#Create a solution vector and populate it
solution <- c()
solution[1] <- round(x1,2)
solution[2] <- round(x2,2)
solution[3] <- round(x3,2)
#return the solution vector
print("The final solution vector")
print(x3)
return (solution)
}m <-matrix(c(1,2,-1,1,-1,-2,3,5,4), ncol=3, nrow=3)
v <- c(1,2,6)
solution <- solve_three_by_three_equations(m,v)## [1] "Below is the original augmented matrix:"
## [,1] [,2] [,3] [,4]
## [1,] 1 1 3 1
## [2,] 2 -1 5 2
## [3,] -1 -2 4 6
##
##
## [1] "The first step applied to the augmented matrix:"
## [,1] [,2] [,3] [,4]
## [1,] 1 1 3 1
## [2,] 0 -3 -1 0
## [3,] -1 -2 4 6
##
##
## [1] "The first column is now complete:"
## [,1] [,2] [,3] [,4]
## [1,] 1 1 3 1
## [2,] 0 -3 -1 0
## [3,] 0 -1 7 7
##
##
## [1] "The second row and column is now complete:"
## [,1] [,2] [,3] [,4]
## [1,] 1 1 3.0000000 1
## [2,] 0 1 0.3333333 0
## [3,] 0 -1 7.0000000 7
##
##
## [1] "The third row now has a 0 in its second column:"
## [,1] [,2] [,3] [,4]
## [1,] 1 1 3.0000000 1
## [2,] 0 1 0.3333333 0
## [3,] 0 0 7.3333333 7
##
##
## [1] "The final matrix:"
## [,1] [,2] [,3] [,4]
## [1,] 1 1 3.0000000 1.0000000
## [2,] 0 1 0.3333333 0.0000000
## [3,] 0 0 1.0000000 0.9545455
##
##
## [1] "The final solution vector"
## [1] 0.9545455solution## [1] -1.55 -0.32 0.95Problem set one was straightforward, by the book. For problem Set two, I struggled a bit to understand the rules which I now understand. The function that I created, solve_three_by_three_equations, works on this particular matrix. I would have to add more code to get it work on different matrices. For example, I would have to add “else” statements for when a coefficient equals 0.