H Solution: Representing Data

H.1 Creating vectors and matrices

# Creating matrices
A1 <- matrix(
  c(1:9),
  nrow = 3
)

A2 <- matrix(
  c(1, 3, 4, 6, 9, 2, 1, 0, 3),
  nrow = 3
)

# Creating vectors
v1 <- c(0, 1, 1)
v2 <- c(2, 1, 0)
v3 <- c(3, 1, 1)
x1 <- c(1, 2, 0, 1)
x2 <- c(2, 3, 1, 1)
x3 <- c(4, 1, 2, 0)

# Another way to create matrices
B1 <- cbind(v1, v3, v2)
B2 <- rbind(v1, v2, v3)

H.2 Solution Q2

Q2: Calculate the following matrices.

• A1 + A2
• A1 * A2
• A1 - A2
• transpose of A1
• transpose of B2

A1 + A2

#>      [,1] [,2] [,3]
#> [1,]    2   10    8
#> [2,]    5   14    8
#> [3,]    7    8   12

A1 * A2

#>      [,1] [,2] [,3]
#> [1,]    1   24    7
#> [2,]    6   45    0
#> [3,]   12   12   27

A1 - A2

#>      [,1] [,2] [,3]
#> [1,]    0   -2    6
#> [2,]   -1   -4    8
#> [3,]   -1    4    6

t(A1)

#>      [,1] [,2] [,3]
#> [1,]    1    2    3
#> [2,]    4    5    6
#> [3,]    7    8    9

t(B2)

#>      v1 v2 v3
#> [1,]  0  2  3
#> [2,]  1  1  1
#> [3,]  1  0  1

H.3 Solution Q3

Q3: Compute the norms of all vectors.

sum(v1^2)

#> [1] 2

sum(v2^2)

#> [1] 5

sum(v3^2)

#> [1] 11

sum(x1^2)

#> [1] 6

sum(x2^2)

#> [1] 15

sum(x3^2)

#> [1] 21

H.4 Solution Q4

Q4: Compute the inverse of A1, A2, and B3. You don’t need to invert the matrices by hand but check they are an inverse.

# Inverse doesn't exist as it is a singular matrix
solve(A1)

#> Error in solve.default(A1): Lapack routine dgesv: system is exactly singular: U[3,3] = 0

solve(A2)

#>            [,1]        [,2]        [,3]
#> [1,] -0.4736842  0.28070175  0.15789474
#> [2,]  0.1578947  0.01754386 -0.05263158
#> [3,]  0.5263158 -0.38596491  0.15789474

solve(A2) %*% A2

#>               [,1]          [,2]          [,3]
#> [1,]  1.000000e+00 -5.551115e-17  2.775558e-17
#> [2,] -2.775558e-17  1.000000e+00 -4.163336e-17
#> [3,] -1.110223e-16  1.665335e-16  1.000000e+00

solve(B2)

#>              v1 v2         v3
#> [1,] -0.3333333  0  0.3333333
#> [2,]  0.6666667  1 -0.6666667
#> [3,]  0.3333333 -1  0.6666667

solve(B2) %*% B2

#>               [,1] [,2] [,3]
#> [1,]  1.000000e+00    0    0
#> [2,]  1.110223e-16    1    0
#> [3,] -1.110223e-16    0    1

H.5 Solution Q5 and Q6

Q5: Compute A1 %*% v1 and v1 %*% A1

A1 %*% v1

#>      [,1]
#> [1,]   11
#> [2,]   13
#> [3,]   15

v1 %*% A1

#>      [,1] [,2] [,3]
#> [1,]    5   11   17

Q6: Compute v2 %*% B2 and B2 %*% v2

v2 %*% B2

#>      [,1] [,2] [,3]
#> [1,]    2    3    2

B2 %*% v2

#>    [,1]
#> v1    1
#> v2    5
#> v3    7