HW1

ASSIGNMENT 1

1. Problem set 1

1. 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)
u%*%v

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

2. 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.

sqrt(u%*%u)

##           [,1]
## [1,] 0.7071068

sqrt(v%*%v)

##      [,1]
## [1,]    5

3. What is the linear combination: 3u − 2v?

(3*u) - (2*v) 

## [1] -4.5  9.5

4. What is the angle between u and v?

#  angle = arccosine(dot product/norm a*norm b )
angle = acos(sum(u*v) / (sqrt(sum(u * u)) * sqrt(sum(v * v))))
round(angle* (180/pi),0)

## [1] 98

2. Problem set 2

Set up a system of equations with 3 variables and 3 constraints and solve for x.

Steps:

1. Upper Triangular Matrix construction

• Start with row 1 of the co-efficient matrix

• Pivot: The first non-zero element in the row being evaluated

• Multiplier: The element being eliminated divided by the Pivot

• Subtract Multiplier times row n from row n+1

• Advance to the next row and repeat

2. Back-substitution to get the solution.

solution <- function(a,b){

# joining matrix and constraint vector
  
mtrx <- cbind(a,b)
  
# assigning a pivot - the first non-zero element of the matrix
pivot1 <- mtrx[1,1]

#  calculating multiplier - the element being eliminated divided by the pivot
multiplier1 <- mtrx[2,1]/pivot1

# subtracting multiplier times row n from row n+1 
mtrx[2,] <- mtrx[2,] - (multiplier1 * mtrx[1,])
  
# advance to the next row and repeat

multiplier2 <- mtrx[3,1]/pivot1
  
mtrx[3,] <- mtrx[3,] - (multiplier2 * mtrx[1,])
  
pivot2 <- mtrx[2,2]
  
multiplier3 <- mtrx[3,2]/pivot2

mtrx[3,] <- mtrx[3,] - (multiplier3 * mtrx[2,])

#  performing back-substitution to get the solution
  
  x3 <- mtrx[3,4]/mtrx[3,3]
  x2 <- (mtrx[2,4] - (mtrx[2,3] * x3))/mtrx[2,2]
  x1 <- (mtrx[1,4] - (mtrx[1,3] * x3) - (mtrx[1,2] * x2))/mtrx[1,1]
  
  return(c(x1, x2, x3))
}

#  testing the function

a<- matrix(c(1,1,3,2,-1,5,-1,-2,4), nrow = 3, ncol = 3, byrow = TRUE)
b <- matrix(c(1,2,6), nrow = 3, ncol = 1, byrow = TRUE)

solution (a, b)

## [1] -1.5454545 -0.3181818  0.9545455