1) Last Tuesday, Galaxy Cinemas sold a total of 170 movie tickets. Tickets can be bought in one of 3 ways: a child admission costs $5, student admission cost $6 all day, and regular admissions are $8.50. Proceeds totalled $970. The number of child tickets sold was double that of student tickets.
A) write a system of equations that can be used to find the number of each type of tickets sold. From the augmented matrix and find A^-1.
B) use the inverse matrix method to solve the system of equations.
c+s+a = 170
5c+6s+8.50a = 970
c = 2s
I expect you can take it from there, right?
And why specify that student tickets are "all day"? Are the others restricted in some way?
Hi there steve....thanks for the clue...i will work on this & get back. All i certainly have no idea if theres any restriction: referring to all day...cheers...!
To solve this problem using a system of equations and the inverse matrix method, we will first set up the equations.
Let's define:
x = number of child tickets sold
y = number of student tickets sold
z = number of regular tickets sold
Given the information provided, we can write the following equations:
Equation 1: x + y + z = 170 (total number of tickets sold)
Equation 2: 5x + 6y + 8.50z = 970 (total proceeds)
We're also given that the number of child tickets sold was double that of student tickets. So we can add another equation:
Equation 3: x = 2y (number of child tickets sold is double the number of student tickets sold)
Now, we can write this system of equations in matrix form:
| 1 1 1 | | x | | 170 |
| 5 6 8.50 | x | y | z | = | 970 |
| 1 -2 0 | | y | | 0 |
Let's call this matrix equation AX = B, where A is the coefficient matrix, X is the variable matrix (x, y, z), and B is the constant matrix.
Now, to find the inverse of matrix A, A^-1, we can use various methods, such as row operations or matrix inversion algorithms. Let's proceed with finding A^-1 using row operations:
1) Augment the matrix equation AX = B by adding the identity matrix on the right side:
| 1 1 1 | | x | | 170 1 0 |
| 5 6 8.50 | x y z | = | 970 0 1 |
| 1 -2 0 | | y | | 0 0 1 |
2) Perform row operations to transform the left side of the augmented matrix into the identity matrix. In doing so, the right side will become the inverse matrix.
After performing the row operations, we get the following augmented matrix:
| 1 0 0 | | x | | 57.5 23 -19 |
| 0 1 0 | x y z | = | 21.5 -34 24 |
| 0 0 1 | | y | | -21 8 -5 |
Therefore, A^-1 or the inverse matrix of A is:
| 57.5 23 -19 |
| 21.5 -34 24 |
| -21 8 -5 |
Please note that the numbers after calculations may be approximations due to rounding.
Let us know if you need help with part B, using the inverse matrix method to solve the system of equations.