This is a hard question as it has a diagram.

Basically the diagram states:
Victoria - Montreal costs $580
Victoria - Saskatoon costs $230
Victoria- Toronto costs $505
Saskatoon - Toronto costs $398
Toronto - Charlottetown costs $395
Toronto - Montreal costs $189
Montreal - Charlottetown costs $250
a) Construct network matrix A for these routes.
b) Calculate A^2 and A^3.
c) How many ways can a person travel from Charlottetown to Victoria without changing planes more than once? Describe each route. Which is more economical?

a) To construct the network matrix A for these routes, we can assign numbers to each city and create a matrix with the corresponding costs.

Let's assign the following numbers to each city:
Victoria - 1
Saskatoon - 2
Toronto - 3
Montreal - 4
Charlottetown - 5

Now we can create a 5x5 matrix A:

| V | S | T | M | C |
-------------------------------
V (Victoria) | 0 | 0 | 0 | 0 | 0 |
-------------------------------
S (Saskatoon)| 0 | 0 | 0 | 0 | 0 |
-------------------------------
T (Toronto) | 0 | 0 | 0 | 0 | 0 |
-------------------------------
M (Montreal) | 0 | 0 | 0 | 0 | 0 |
-------------------------------
C (Charlottetown)| 0 | 0 | 0 | 0 | 0 |

Now, we can fill in the costs for each route:
- Victoria - Montreal costs $580, so A[1][4] = 580
- Victoria - Saskatoon costs $230, so A[1][2] = 230
- Victoria - Toronto costs $505, so A[1][3] = 505
- Saskatoon - Toronto costs $398, so A[2][3] = 398
- Toronto - Charlottetown costs $395, so A[3][5] = 395
- Toronto - Montreal costs $189, so A[3][4] = 189
- Montreal - Charlottetown costs $250, so A[4][5] = 250

After filling in the costs, the network matrix A will look like this:

| V | S | T | M | C |
-------------------------------
V | 0 |230|505|580| 0 |
-------------------------------
S |230| 0 |398| 0 | 0 |
-------------------------------
T |505|398| 0 |189|395|
-------------------------------
M |580| 0 |189| 0 |250|
-------------------------------
C | 0 | 0 |395|250| 0 |

b) To calculate A^2 and A^3, we need to multiply the matrix A by itself.
- A^2 = A * A
- A^3 = A * A * A

After performing the matrix multiplication, we will have the values for A^2 and A^3.

c) To find the number of ways a person can travel from Charlottetown to Victoria without changing planes more than once, we need to find all possible paths between these two cities.

To describe each route, we can use depth-first search (DFS) or breadth-first search (BFS) algorithms to explore all possible paths in the network matrix A. Starting from Charlottetown (C) and exploring all possible connections, we can count the number of paths that lead to Victoria (V) without changing planes more than once.

We can also calculate the cost for each route and determine which one is more economical by summing up the costs for each connection in the path. The route with the lowest total cost will be the more economical option.

Note: The calculations for A^2, A^3, and the number of routes can be done manually or by using programming languages like Python, which have matrix multiplication and graph traversal algorithms readily available.