Problem set 1

  1. Calculate the dot product u.v where u = [0.5; 0.5] and v = [3; -4]

Sol:

u <- c(0.5, 0.5)
v <- c(3, -4)
 The dot product is -0.5.
  1. 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.

Sol:

The length of u is 0.7071068 and length of v is 5.
  1. What is the linear combination: 3u - 2v?

Sol:

The linear combination is -4.5, 9.5.
  1. What is the angle between u and v

Sol: The angle between u and v is 1.7126934

Problem set 2

Set 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.

Please test it with the system below and it should produce a solution x = [-1.55, -0.32, 0.95]

\[ \begin{bmatrix} 1 & 1 & 3 \\ 2 & -1 & 5 \\ -1 & -2 & 4 \end{bmatrix}\, \begin{bmatrix} x_1 \\x_2 \\x_3 \end{bmatrix}= \begin{bmatrix} 1 \\2 \\6 \end{bmatrix} \]

Solution to problem set 2

Create Function

sol <- function(A, b){
 #Pivot
  pivot <- A[1,1]
  #exchange rows where pivot = 0
  if (pivot == 0){
    A <- A[c(2,1,3),]
  pivot <- A[1,1]
  if (pivot == 0){
    A <- A[c(3,2,1),]
  pivot <- A[1,1]
  }
  }
  #Multiplier
  mplyr_1 <- A[2,1]/pivot
  mplyr_2 <- A[3,1]/pivot
  #Elimination: row2
  A[2, ] <- A[2, ] - mplyr_1 * A[1, ]
  b[2,] <- b[2,] - mplyr_1 * b[1,]
  #Elimination: row3
    A[3, ] <- A[3, ] - mplyr_2 * A[1, ]
  b[3,] <- b[3,] - mplyr_2 * b[1,]
  
  #Repeat
  
  pivot_1 <- A[2,2]
  #exchange rows where pivot = 0
  if (pivot_1 == 0){
    A <- A[c(1,3,2),]
  pivot_1 <- A[2,2]
  }
  #Multiplier
  mplyr_3 <- A[3,2]/pivot_1
  #Elimination: row3
  A[3, ] <- A[3, ] - mplyr_3 * A[2, ]
  b[3,] <- b[3,] - mplyr_3 * b[2,]
  
  #x3 solution
  x3 <- b[3] / A[3, 3]
  
  #substitution1
  x2 <- (b[2] - A[2, 3] * x3) / A[2, 2]
  #substitution2
  x1 <- (b[1] - A[1, 3] * x3 - A[1, 2] * x2) / A[1, 1]
  
  #results
  x <- matrix(c(x1, x2, x3), nrow = 3)
  x
  
}

Test Function

A <- array(c(1, 2, -1, 1, -1, -2, 3, 5, 4), dim = c(3,3))
b <- array(c(1, 2, 6), dim = c(3,1)) 
round(sol(A, b), 2)
##       [,1]
## [1,] -1.55
## [2,] -0.32
## [3,]  0.95