please look at the attached final answer. in case there is need for review, I am readily available.

Surname: 1
Name:
Instructor’s name:
Course:
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Question 1
We describe a permutation matrix as a square binary matrix that has exactly one entry of 1 in
each column row and each row and 0s elsewhere
a) the product of permutation matrices is a permutation matrix as shown below
0
[
1

(0 ∗ 0) + (1 ∗ 1)
1
0 1
]∗[
]=[
(1 ∗ 0) + (0 ∗ 1)
0
1 0

(0 ∗ 1) + (1 ∗ 0)
0 1
]=[
]
(1 ∗ 1) + (0 ∗ 0)
1 0

a) the inverse of a permutation matrix is the transpose of the matrix as shown here:
0
[
1

=[

1
𝑎
]∗[
0
𝑐

1
𝑏
]=[
0
𝑑

0
]
1

(0 ∗ 𝑎) + (1 ∗ 𝑏) (0 ∗ 𝑏) + (1 ∗ 𝑑)
𝑏
]=[
(1 ∗ 𝑎) + (0 ∗ 𝑐) (1 ∗ 𝑏) + (0 ∗ 𝑑)
𝑎

0
Therefore b=1, d=0, a=0, c=1 which is the transpose of [
1

1
𝑑
]=[
0
𝑐

0
]
1

1
]
0

Question 2
We re given that that g, r, and C are known.
𝑟

The first thing is to make x the subject of the formula g’ x = r to obtain x=𝑔′
Also, the condition that the portfolio weights sum to one can be expressed as

Where 1 is a 3 × 1 vector with each element equal to 1. Consider another portfolio with weights
y = (y y yc)’,

Surname: 2

The first three elements of z
 are the portfolio weights m = (  )0 for the global minimum variance portfolio with
expected return P,M = m0μ and variance 2
 = m0Σm
Question 3
A linear system with unique solution has a solution set with one element. A linear system
with no solutions has an empty set of solutions
Ax=0 is an homogeneous equation since it has a constant ...