Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
2. Matrix Multiplication
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Exercise 28 Page 777

Practice makes perfect
a We are told that a florist makes three types of special floral arrangements. Let's start by organizing the given information in a table.
Flower
Floral Arrangement Number of lilies Number of daisies Number of carnations
First 3 0 0
Second 3 0 4
Third 0 4 3

We can write this table as a matrix A. A= 3 & 0 & 0 3 & 0 & 4 0 & 4 & 3 Please note that this is not the only answer, as the type of flower can be arranged as we want. We only have to be careful and follow this choice for the remainder of the exercise.

b We are told that lilies cost $2.15 each, carnations cost $0.90 each, and daisies cost $1.30 each. We can arrange this information in a 3 * 1 matrix P. Let's be careful and follow the same ordering that we used in Part A!

P= 2.15 1.30 0.90 l← Price of each lily ← Price of each daisy ← Price of each carnation

Please note that this matrix depends on the ordering used in Part A.
c Since the number of flowers of each arrangement are arranged in the rows of A and their prices are in the column of P, we can use matrix multiplication to find the total prices.
The matrix AP will be a 3* 1 matrix with the total price of each arrangement. AP= 3 & 0 & 0 3 & 0 & 4 0 & 4 & 3 2.15 1.30 0.90 Let's calculate it!
AP= 3 & 0 & 0 3& 0 & 4 0 & 4 & 3 2.15 1.30 0.90
AP= 3(2.15) + 0(1.30) + 0(0.90) 3(2.15) + 0(1.30) + 4(0.90) 0(2.15) + 4(1.30) + 3(0.90)
AP= 6.45+0+0 6.45+0+3.60 0+5.20+2.70
AP= 6.45 10.05 7.90
We found the matrix AP with the prices of each floral arrangement. AP= 6.45 10.05 7.90 l← Price of first arrangement ← Price of second arrangement ← Price of third arrangement