958 CHAPTER 9 Systems and Matrices An Application of Matrix Algebra EXAMPLE 8 Using Matrix Multiplication to Model Plans for a Subdivision A contractor builds three kinds of houses, models A, B, and C, with a choice of two styles, colonial or ranch. Matrix P below shows the number of each kind of house the contractor is planning to build for a new 100-home subdivision. The amounts for each of the main materials used depend on the style of the house. These amounts are shown in matrix Q, while matrix R gives the cost in dollars for each kind of material. Concrete is measured here in cubic yards, lumber in 1000 board feet, brick in 1000s, and shingles in 100 square feet. Colonial Ranch Model A Model B Model C £ 0 30 10 20 20 20§ = P Concrete Lumber Brick Shingles Cost per Unit Colonial Ranch c 10 2 0 2 50 1 20 2d = Q Concrete Lumber Brick Shingles ≥ 20 180 60 25¥ = R (a) What is the total cost of materials for all houses of each model? (b) How much of each of the four kinds of material must be ordered? (c) What is the total cost of the materials? SOLUTION (a) To find the materials cost for each model, first find matrix PQ, which will show the total amount of each material needed for all houses of each model. Concrete Lumber Brick Shingles PQ= £ 0 10 20 30 20 20§ c 10 50 2 1 0 20 2 2d = £ 1500 30 600 60 1100 40 400 60 1200 60 400 80§ Model A Model B Model C Multiplying PQ and the cost matrix R gives the total cost of materials for each model. Cost 1PQ2R = £ 1500 1100 1200 30 40 60 600 400 400 60 60 80§ ≥ 20 180 60 25¥ = £ 72,900 54,700 60,800§ Model A Model B Model C (b) To find how much of each kind of material to order, refer to the columns of matrix PQ. The sums of the elements of the columns will give a matrix whose elements represent the total amounts of all materials needed for the subdivision. Call this matrix T, and write it as a row matrix. T = 33800 130 1400 2004
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