The top row of matrix A (1, 2, 1) is multiplied with the first column of matrix B (1,0,-1) and the result is 1x1 + 2x0 + 1x -1 = 0 this is row 1 column 1 of the resultant matrix
The top row of matrix A (1,2,1) is multiplied with the second column of matrix B (-1, -1, 1) and the result is 1 x-1 + 2 x -1 + 1 x 1 = -2 , this is row 1 column 2 of the resultant matrix
Repeat with the second row of matrix A (-1,-1.-2) x (1,0,-1) = 1 this is row 2 column 1 of the resultant matrix, multiply the second row of A (-1,-1,-2) x (-1,-1,1) = 0, this is row 2 column 2 of the resultant
Repeat with the third row of matrix A( -1,1,-2) x (1,0, -1) = 1, this is row 3 column 1 of the resultant
the third row of A (-1,1,-2) x( -1,-1,1) = -2, this is row 3 column 2 of the resultant matrix
Matrix AB ( 0,-2/1,0/1,-2)
Given:
The cost of adults ticket = $18
The cost of children's ticket = $8.25
Total tickets = 2300
Total revenue = $30,168.
To find:
The number of children and number of adults attended the zoo that day.
Solution:
Let x be the number of children and y be the number of adults.
Equation for tickets:
...(i)
Equation for revenue:
...(ii)
Plot the graphs of the given equations on a coordinate plane as shown below.
From the graph it is clear that the graph of both equations intersect each other at (1148,1152).
It means the number of adults is 1148 and the number of children is 1152.
It can be solved algebraically as shown below:
Substitute the value of y in (ii) from (i).




Divide both sides by 9.75.


Putting
in (i), we get



Therefore, the number of adults is 1148 and the number of children is 1152.
Answer:
If the line is horizontal, then all that have a 2 in the Y space will apply because the line goes across meaning that it won't change on the vertical axis.
(5,2) and (-2, 2)