To solve this problem using matrices, you can set up a system of equations based on the rental and return patterns provided.
Let's denote the number of cars at each location at the start of the week using a column vector, C = [A, B, C]. Similarly, let R = [A, B] represent the number of cars returned to each location at the end of the week.
Based on the patterns given, we can write the following equations:
For location A:
- The number of cars at A at the end of the week is the sum of those returned from A and B: R_A = (17/20)C_A + (1/10)C_B
- The number of cars at A at the start of the next week should be the same as at the start of the week: C_A = C_A
For location B:
- The number of cars at B at the end of the week is the sum of those returned from B and A: R_B = (4/5)C_B + (1/10)C_A
- The number of cars at B at the start of the next week should be the same as at the start of the week: C_B = C_B
For location C:
- The number of cars at C at the end of the week is the sum of those returned from A and B: R_C = (1/10)C_A + (1/20)C_B
- The number of cars at C at the start of the next week should be the same as at the start of the week: C_C = C_C
Now, we can rewrite these equations in matrix form as:
[R_A] [17/20 1/10 0] [C_A]
[R_B] = [1/10 4/5 0] * [C_B]
[R_C] [1/10 0 1/20] [C_C]
To find the values of C_A, C_B, and C_C, you can use matrix algebra to solve this system of equations. Let me do the calculations for you:
[R_A] [17/20 1/10 0]^-1 [C_A]
[R_B] = [1/10 4/5 0] * [C_B]
[R_C] [1/10 0 1/20] [C_C]
Here, ^-1 represents the inverse of the matrix [17/20 1/10 0].
By multiplying both sides of the equation by the inverse matrix, we can isolate [C_A, C_B, C_C] on one side:
[C_A] = [17/20 1/10 0]^-1 * [R_A]
[C_B] [1/10 4/5 0] [R_B]
[C_C] [1/10 0 1/20] [R_C]
Now, substitute the given values of [R_A, R_B, R_C] into the equation and perform the matrix multiplication to find the values of [C_A, C_B, C_C].
I hope this helps you get started with solving the problem using matrices. Let me know if you have any further questions!