Applied Linear Algebra: Equilibrium Prices and Sector Analysis

This assignment solution addresses the problem of determining equilibrium prices in a two-sector economy using applied linear algebra. The solution begins by applying Polya's four-step problem-solving method, starting with understanding the problem of determining the equilibrium prices of goods and services. It then constructs an exchange table to represent the flow of goods and services between the two sectors, with Goods selling 80% of its output to Services, and Services selling 70% to Goods. The solution uses the exchange table to calculate the income and expenses for each sector, forming a system of linear equations. The equations are then represented in matrix form and solved, including row reduction of the augmented matrix. The general solution is derived, and the equilibrium prices are determined. The solution emphasizes that only the price ratios matter and that the equilibrium price is unaffected by proportionate price changes. The assignment demonstrates the application of linear algebra principles to economic modeling.

Running head: APPLIED LINEAR ALGEBRA 1
Applied Linear Algebra
Name
Institution
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APPLIED LINEAR ALGEBRA 2
Applied Linear Algebra
Question 1
According to Polya, four step-by-step techniques are used to solve any problem. First,
understand the problem.
Evidently, the problem requires the equilibrium prices of two sectors (goods and services) that
can make expenditures equivalent to expenses. In the question, Goods sell 80% worth of its
outputs to the services economy. On the other hand, Services sell 70% worth of outputs to Goods
sector.
Secondly, we devise a plan to solve the problem. The plan involves constructing an exchange
table. Then, fill in the columns of the exchange table. After that, we denote the output of the
Goods and Services as MG and MS respectively.
We then carry out the plan as follows:
The exchange table is shown in table 1, where the column entries define the destination of a
sector’s output.
Notably, Goods retain (100-80)/100=0.2 whereas Services retain (100-70)/100=0.3
Dispersal of outputs from a sector
Sector Output Goods (MG) Services (MS) Sector input Purchaser
↓ ↓
0.20 0.70 → Goods
0.80 0.30 → Services
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APPLIED LINEAR ALGEBRA 3
Then, denoting the output of the Goods and Services as MG and MS respectively and determining
the total input to the goods in the first row we obtain, 0.20 M G +0.70 M S. Therefore, the income
for goods is MG whereas the expenses are 0.80 M G +0.30 MS. Similarly, from row 2 the input or
rather the expenses will be given as 0.80 MG +0.30 MS and the income as MS as shown in table 2.
Sector income Sector expenses
MG = 0.20 M G +0.70 M S
MS =0.80 M G +0.30 M S
M G=0.20 M G +0.70 M S
M S =0.80 MG +0.30 M S
Moving the variables MG and MS to the left and putting the like terms together from the above
equations:
0.80 MG−0.70 M S =0
−0.80 MG +0.70 MS =0
Forming a matrix and then “row reduce the augmented matrix,”
[ 0.80 −0.70 0.00
−0.80 0.70 0.00 ] → [ 0.80 −0.70 0.00
0.00 0.00 0.00 ] → [ 1 −0.875 0.00
0.00 0.00 0.00 ]
Therefore, M G=0.875∧M S =1
Then, the general solution can be denoted as M G=0.875 MS. But we know that the natural prices
can be represented in a whole number. Thus, the equilibrium prices will be M G=875 and
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APPLIED LINEAR ALGEBRA 4
M S =1000 or M G=70 and M S =80 (from the relation M G=0.875 M S ¿ . Evidently, it is only the
price ratios that matter: M G=0.875 M S . The equilibrium price is not affected by proportionate
price changes.
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