Calculate the dot product u.v where u = [0.5; 0.5] and v = [3; −4]
u <- c(0.5, 0.5)
v <- c(3, -4)
print(paste("The dot product of u, v is:", u %*% v))## [1] "The dot product of u, v is: -0.5"What are the lengths of u and v? Please note that the mathematical notion of the length of a vector is not the same as a computer science definition.
The length of a vector is given by the euclidean distance or by use of the Pythagoras therm.
The formula for this is \(x^2 + y^2 = z^2\) as such:
LenU <- sqrt(sum(u^2))
LenV <- sqrt(sum(v^2))
print(paste('The length of u is ', LenU, 'and the length of v is', LenV))## [1] "The length of u is 0.707106781186548 and the length of v is 5"What is the linear combination: 3u − 2v?
linComb <- 3 * u - 2 * v
print(paste(c('The linear combination of the vectors and constants is:', linComb), collapse = " "))## [1] "The linear combination of the vectors and constants is: -4.5 9.5"What is the angle between u and v?
The angel can be given with the \(cosin\) distance.
Special thanks to this stackoverflow question.
theta <- acos( sum(u*v) / ( sqrt(sum(u * u)) * sqrt(sum(v * v))))
theta * 180 / pi## [1] 98.1301Set 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. You don’t have to worry about degenerate cases and can safely assume that the function will only be tested with a system of equations that has a solution. Please note that you do have to worry about zero pivots, though. Please note that you should not use the built-in function solve to solve this system or use matrix inverses. 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. 12 IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS - FALL 2014 Please test it with the system below and it should produce a solution x = [−1.55, −0.32, 0.95]
\[\mathbf{} \left[\begin{array} {rrr} 1 & 1 & 3 \\ 2 & -1 & 5 \\ -1 & -2 & 4 \end{array}\right] % \left[ \begin{array}{ccc} 1 \\ 2 \\ 6 \end{array} \right] \]
Please send your code (as an R-markdown file, named using your first initial, last name, assignment, problem set. For instance, if I submit the code for this assignment, it will be called GIyengar Assign1.Rmd
Instead of writing out a function that would do the exact steps I wanted to use loops to make my code shorter. I went down quite the rabbit hole trying to make it work with for loops but this was exceedingly difficult.
I stumbled upon this that would work on any n dimensional matrix. I modified it to work in a function. The result as shown below the initial solution works on matrixies of any size.
matrixA <- matrix(
c(1,1,3,2,-1,5,-1,-2,4),
nrow=3,
byrow = TRUE)
vectorb <- matrix(
c(1,2,6),
ncol=1,
byrow = TRUE)loopFunc <- function(A, b){
p <- nrow(A)
aug <- cbind(A, b)
aug[1, ] <- aug[1, ] / aug[1,1] # Normaliz the fist number to 1. This makes everything easier.
i <- 2 # So that we don't run the code on an index out of range e.g., A[1-1, ]
while (i < p + 1) {
j <- i # This effectively sets the loop to go over one set and then the other.
while (j < p + 1) {
aug[j, ] <- aug[j, ] - aug[i -1, ] * aug[j, i-1] # The last argument here works because we normalized the first row to 1
# The above command also sets the jth row of the matrix to the correct result.
j <- j + 1 # This makes the code easier than in a for loop.
}
while (aug[i, i] == 0) {
aug <- rbind(aug[-1], aug[i]) # Checks to make sure that we don't have zero devision
}
aug[i, ] <- aug[i, ] / aug[i, i] # Scales the row to 1 again, making calculations easier.
i <- i + 1
}
for (i in p:2) {
# p is now a large number and we need to go in reverse to solve the upper triangle.
for (j in i:2-1) { # The same logic holds here but for the second element.
aug[j, ] <- aug[j, ] - aug[i, ] * aug[j, i] # Solves the rows going up.
}
}
return(aug)
}
mx <- loopFunc(matrixA, vectorb)
mx## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 -1.5454545
## [2,] 0 1 0 -0.3181818
## [3,] 0 0 1 0.9545455set.seed(101)
A <- matrix(floor(runif(100, 1,11)) , 10)
b = floor(runif(10, 1 , 11))
A## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 4 9 8 5 5 1 9 3 7 7
## [2,] 1 8 10 4 5 5 9 10 8 10
## [3,] 8 8 3 2 8 7 7 6 8 8
## [4,] 7 10 7 2 4 8 10 9 10 2
## [5,] 3 5 10 6 4 5 9 9 1 3
## [6,] 4 6 8 10 3 8 8 3 10 9
## [7,] 6 9 1 3 1 7 1 1 4 4
## [8,] 4 3 4 9 10 6 4 3 5 3
## [9,] 7 5 5 1 8 7 9 7 2 1
## [10,] 6 1 7 10 2 6 10 10 1 1b## [1] 2 1 4 9 8 4 1 1 8 10loopFunc(A, b)## b
## [1,] 1 0 0 0 0 0 0 0 0 0 2.81105015
## [2,] 0 1 0 0 0 0 0 0 0 0 -1.51843410
## [3,] 0 0 1 0 0 0 0 0 0 0 2.87764776
## [4,] 0 0 0 1 0 0 0 0 0 0 -0.85211352
## [5,] 0 0 0 0 1 0 0 0 0 0 -0.38072851
## [6,] 0 0 0 0 0 1 0 0 0 0 -0.23628663
## [7,] 0 0 0 0 0 0 1 0 0 0 -1.57768373
## [8,] 0 0 0 0 0 0 0 1 0 0 0.07313526
## [9,] 0 0 0 0 0 0 0 0 1 0 0.51133983
## [10,] 0 0 0 0 0 0 0 0 0 1 -0.25694428