Assignment-1 :: Fundamentals of Computational Mathematics

---

Problem Set 1:

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

u <-  c(0.5,0.5)
v <- c(3,-4)

dot.product<-u%*%v
dot.product

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

(2) What are the lengths of \(u\) and \(v\)?

Length of u:

Length_of_u<-sqrt(u %*% u)
Length_of_u

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

Length of v:

Length_of_v<-sqrt(v %*% v)
Length_of_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

theta <- acos(dot.product/(Length_of_u*Length_of_v))
theta

##          [,1]
## [1,] 1.712693

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 \(x = [-1.55;-0.32;0.95]\) \[\begin{bmatrix}1 & 1 & 3 \\2 & -1 & 5 \\-1 & -2 & 4\end{bmatrix}\,\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=\begin{bmatrix}1 \\2 \\6\end{bmatrix}\]

# from http://stackoverflow.com/questions/16044377/how-to-do-gaussian-elimination-in-r-do-not-use-solve

A <- array(c(1, 2, -1, 1, -1, -2, 3, 5, 4), dim=c(3,3))
b <- c(1, 2, 6) # constraint vector

solve_this<-function (A,b)
{
  p <- nrow(A)
  
  # binding A and b
  binded <- cbind(A,b)  
  binded[1,] <- binded[1,]/binded[1,1]
  
  for (i in 2:p)
  {
       for (j in i:p) 
       {
            binded[j, ] <- binded[j, ] - binded[i-1, ] * binded[j, i-1]
       }
       binded[i,] <- binded[i,]/binded[i,i]
  }
  
  for (i in p:2)
  {
       for (j in i:2-1) 
       {
            binded[j, ] <- binded[j, ] - binded[i, ] * binded[j, i]
       }
  }
  print (binded)
}

solve_this(A,b)

##                     b
## [1,] 1 0 0 -1.5454545
## [2,] 0 1 0 -0.3181818
## [3,] 0 0 1  0.9545455

Verifying using solve function:

A <- array(c(1, 2, -1, 1, -1, -2, 3, 5, 4), dim=c(3,3))
b <- c(1, 2, 6)
solve(A, b)

## [1] -1.5454545 -0.3181818  0.9545455