1. Problem set 1 You can think of vectors representing many dimensions of related information. For instance, Netflix might store all the ratings a user gives to movies in a vector. This is clearly a vector of very large dimensions (in the millions) and very sparse as the user might have rated only a few movies. Similarly, Amazon might store the items purchased by a user in a vector, with each slot or dimension representing a unique product and the value of the slot, the number of such items the user bought. One task that is frequently done in these settings is to find similarities between users. And, we can use dot-product between vectors to do just that. As you know, the dot-product is proportional to the length of two vectors and to the angle between them. In fact, the dot-product between two vectors, normalized by their lengths is called as the cosine distance and is frequently used in recommendation engines.
library('geometry')
## Loading required package: magic
## Warning: package 'magic' was built under R version 3.4.3
## Loading required package: abind
  1. Calculate the dot product u.v where u = [0.5; 0.5] and v = [3; βˆ’4]
u <- c(0.5, 0.5)
v <- c(3,-4)
dotpro <- u%*%v
dotpro
##      [,1]
## [1,] -0.5
paste0("dot product is: ", dotpro)
## [1] "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.

Length of vector is usually given by formula v = √(x^2 + y^2)) Calculated manually using above formula U = sqrt [ (0.5)^2 + (0.5)^2] = sqrt(0.5) = 0.71 V = sqrt [ (3)^2 + (-4)^2 ] = sqrt (25) = 5

length(u)
## [1] 2
length(v)
## [1] 2
  1. What is the linear combination: 3u βˆ’ 2v?
lin_com <- 3*u - 2*v
paste0("linear combination is", lin_com)
## [1] "linear combination is-4.5" "linear combination is9.5"
  1. What is the angle between u and v

To calculate angle we have to use formula Cos πžͺ = uβ‹…v/ β€–uβ€–β‹…β€–vβ€– i.e dot product (uv) / length of u * length of v

angle <- acos(dot(u,v) / (0.71*5))
angle1 <- angle * 180/pi   #to convert radian into degrees 
angle1
## [1] 98.09675

You can use R-markdown to submit your responses to this problem set. If you decide to do it in paper, then please either scan it or take a picture using a smartphone and attach that picture. Please make sure that the picture is legible before submitting.

  1. 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 = [βˆ’1.55, βˆ’0.32, 0.95]

Solving using the pivot and multiply (1) Start with row 1 of the co-efficient matrix (2) Pivot: The first non-zero element in the row being evaluated (3) Multiplier: The element being eliminated divided by the Pivot (4) Subtract Multiplier times row n from row n+1 (5) Advance to the next row and repeat

solve_equation <- function(a, b) {
    

  a <- cbind(a, b)
  pivot <- a[1, 1]
  multi <- a[2, 1]/pivot
  row1 <- a[1,]
  row2 <- a[2,]
  a[2,] <- row2 - row1 * multi
  
  
  pivot <- a[1, 1]
  multi <- a[3, 1]/pivot
  row1 <- a[1,]
  row3 <- a[3,]
  a[3,] <- row3 - row1 * multi
  
  
  pivot <- a[2, 2]
  multi <- a[3, 2]/pivot
  row2 <- a[2,]
  row3 <- a[3,]
  a[3,] <- row3 - row2 * multi
  
  
  x3 <- a[3, 4] / a[3, 3]
  x2 <- (a[2, 4] - a[2, 3] * x3) / a[2, 2]
  x1 <- (a[1, 4] - a[1, 3] * x3 - a[1, 2] * x2) / a[1, 1]
  value <- c(x1, x2, x3)
  return(round(value, 2))
}

matrixA <- c(1, 2, -1, 1, -1, -2, 3, 5, 4)
constantB <- c(1, 2, 6)
FinalValues <- matrix(matrixA, 3, 3)

solve_equation(FinalValues, constantB)
##     b     b     b 
## -1.55 -0.32  0.95