McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
8. Solving Systems of Equations Using Inverse Matrices
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Exercise 36 Page 203

Practice makes perfect
a Let's use a table to organize the information given in the question about the movements.
From
City Suburbs
To City 95 % 3 %
Suburbs 5 % 97 %

We can write a transition matrix corresponding to this table. 0.95& 0.03 0.05& 0.97

b Let's see why the transition matrix of Part A is useful. To do this, let's use a 2* 1 matrix for the population of the city (c) and the suburbs (s) at any given point.

c s Let's calculate and interpret the product matrix. 0.95& 0.03 0.05& 0.97 c s = 0.95c+0.03s 0.05c+0.97sThe result is a 2* 1 matrix, and the elements are the sum of two products.

  • The expression 0.95c gives the number of people who stayed in the city.
  • The expression 0.03s gives the number of people who moved from the suburbs to the city.
  • The sum 0.95c+0.03s is therefore the population of the city in the next year. This is the first element of the product matrix.
  • Similarly, 0.05c+0.97s is the population of the suburbs in the next year. This is the second element of the product matrix.

In this part of the question we are given the current population of the city and the suburbs. c&=16 275 s&=17 552 We are asked to predict the population next year. According to our interpretation of the product above, this can be done by calculating the following product matrix. 0.95& 0.03 0.05& 0.97 16 275 17 552 Let's evaluate the product.

0.95& 0.03 0.05& 0.97 16 275 17 552
0.95(16 275)+0.03(17 552) 0.05(16 275)+0.97(17 552)
15 461.25+526.56 813.75+17 025.44
15 987.81 17 839.19

Since the second element of the result corresponds to the population in the suburbs, according to this trend there will be about 17 839 people living in the suburbs next year.

c According to the interpretation of the matrix product in Part B, we can find the population in the previous year by solving the following matrix equation.

0.95& 0.03 0.05& 0.97 c s = 16 275 17 552 A matrix equation like this can be solved using the inverse matrix. MX=P ⟹ X=M^(-1)P We will use the determinant to find the inverse of a 2* 2 matrix M= 0.95& 0.03 0.05& 0.97 ⇓ |M|=0.95(0.97)-0.03(0.05)=0.92 ⇓ M^(-1)=1/0.92 0.97& -0.03 -0.05& 0.95 Let's use this inverse to find the answer to the question.

c s =M^(-1)P
c s =1/0.92 0.97& -0.03 -0.05& 0.95 16 275 17 552
â–¼
Evaluate right-hand side
c s =1/0.92 0.97(16 275)+(-0.03)(17 552) -0.05(16 275)+0.95(17 552)
c s =1/0.92 15 786.75+(-526.56) -813.75+16 674.40
c s =1/0.92 15 260.19 15 860.65

Multiply matrix by 1/0.92

c s = 16 587.16 17 239.84

Since the first element of the result corresponds to the population in the city, we found that there were about 16 587 people living in the city last year.