CSmith_Assign1

PROBLEM SET 1

u <- c(.5,.5)
v <- c(3,-4)

# 1
dotproduct <- u%*%v
dotproduct

##      [,1]
## [1,] -0.5

#-0.5

# 2
length_u <- sqrt((.5)^2+(.5)^2)
length_u

## [1] 0.7071068

#0.707107

# 
length_v <- sqrt((3)^2+(-4)^2)
length_v

## [1] 5

#5

# 3
comb <- 3*u + 2*v
comb

## [1]  7.5 -6.5

#(7.5, -6,5)

# 4
theta <- (dotproduct)/(sqrt(.5^2+.5^2)+sqrt(3^2+(-4)^2))
degrees <- acos(theta) * (180/pi)
degrees

##          [,1]
## [1,] 95.02613

#95.02613 degrees

PROBLEM SET 2

#Matrix and vector that will be entered as variables into the function
matrixA <- matrix(c(1,2,-1,1,-1,-2,3,5,4), ncol = 3)
vectorB <- c(1,2,6)


upperTriangleSolution <- function(matrixA, vectorB)
{
  #Check for Zero Pivot
  #If a row has a leading zero, it will swap with a non-leading zero row
  #If the first column is all rows, the function will end
  if(matrixA[1,1] != 0){
    result <- "TRUE"
  }else
  {
    if(matrixA[1,1] == 0 & matrixA[2,1] != 0)
    {
      tempRow <- matrixA[2,]
      matrixA[2,] <- matrixA[1,]
      matrixA[1,] <- tempRow
      result <- "TRUE"
    }else if(matrixA[1,1] == 0 & matrixA[3,1] != 0)
    {
      tempRow <- matrixA[3,]
      matrixA[3,] <- matrixA[1,]
      matrixA[1,] <- tempRow
      result <- "TRUE"
    }else{result <- "FALSE"}
  }
  
  #If there is a non-zero in the first column, the function
  #will construct an upper triangle matrix
  if(result == "TRUE")
  {
    #Construct Upper Triangle Matrix by finding the coefficient
    #to multiply by the first row and adding to the second row
    #The first two numbers in column one are compared
    a <- matrixA[1,1]
    b <- matrixA[2,1]
    getZero <- -(b/a)
    matrixA[2,] <- (matrixA[1,] * getZero) + matrixA[2,]
    
    #Comparing first and third numbers in the first column 
    b <- matrixA[3,1]
    getZero <- -(b/a)
    matrixA[3,] <- (matrixA[1,] * getZero) + matrixA[3,]
    
    #At this point only the first number in the first column is a non-zero
    #The second and third numbers of the second column are compared, multiplied, and added
    a <- matrixA[2,2]
    b <- matrixA[3,2]
    getZero <- -(b/a)
    matrixA[3,] <- (matrixA[2,] * getZero) + matrixA[3,]
    
    #Backwards Substitution to solve for X variables
    getX3 <- vectorB[3]/matrixA[3,3]
    
    getX2 <- (vectorB[2]-(matrixA[2,3]*getX3))/matrixA[2,2]
    
    getX1 <- (vectorB[1]-(matrixA[1,2]*getX2)-(matrixA[1,3]*getX3))/matrixA[1,1]
    
    print("Upper Triangle Matrix is: ")
    print(matrixA)
    
    message("X1 = ", getX1)
    message("X2 = ", getX2)
    message("X3 = ", getX3)
    
  }else
  {
    print("Can't be computed.  1st column is all zeros")
    print(matrixA)
  }
}