An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: Modeling an Economy

An economy is a very complex system. An economy is not a well-defined object, there are many factors that influence an economy, and it is often unclear how the factors influence each other. Mathematical modeling plays an important role in attempting to understand an economy.

In 1941 Wassily Leontief developed the first empirical model of a national economy. Around 1949 Leontief used data from the U.S. Bureau of Labor Statistics to divide the U.S. economy into 500 sectors. He then set up linear equations for each sector. This system was too large for the computers at the time to solve, so he then aggregated the information into 42 sectors. The Harvard Mark II computer was used to solve this system, one of the first significant uses of computers for mathematical modeling. Leontief won the 1973 Nobel Prize in economics for his work.

With such large models (Leontief's models are called input-output models) it is important to find a shorthand way to represent the resulting systems. In this section we will see how to represent any size system of linear equations in a very convenient way. Later, we will analyze a small economy using input-output models.

Subsection Introduction

There is another useful way to represent a system of linear equations using a matrix-vector product that we investigate in this section. To understand how this product comes about, recall that we can represent the linear system

as a vector equation as

We can view the left hand side of Equation (5.1) as a matrix-vector product. Specifically, if \(A = \left[ \begin{array}{crrr} 1\amp 4\amp 2\amp 4 \\ 2\amp -1\amp -5\amp -1 \\ 3\amp 7\amp 1\amp 7 \end{array} \right]\) and \(\vx = \left[ \begin{array}{c} x_1 \\x_2 \\x_3 \\x_4 \end{array} \right]\text{,}\) then we define the matrix-vector product \(A\vx\) as the left hand side Equation (5.1). So the matrix-vector product \(A\vx\) is the linear combination of the columns of \(A\) with weights from the vector \(\vx\) in order.

With this definition, the vector equation in (5.1) can be expressed as a matrix-vector equation as

We call this representation the matrix-vector form of the system. Note that the matrix \(A\) in this expression is the same as the coefficient matrix that appears in the augmented matrix representation of the system.

We can use the above definition of the matrix-vector product as a linear combination with any matrix and any vector, as long as it is meaningful to use the entries in the vector as weights for the columns of the matrix. For example, for \(A=\left[ \begin{array}{cc} 1\amp 2\\3\amp 1\\1\amp 1 \end{array} \right]\) and \(\vv = \left[ \begin{array}{c} 3\\4 \end{array} \right]\text{,}\) then we can define \(A\vv\) to be the linear combination of the columns of \(A\) with weights 3 and 4:

However, note that if \(\vv\) had three entries, this definition would not make sense since we do not have three columns in \(A\text{.}\) In those cases, we say \(A\vv\) is not defined. We will later see that this definition can be generalized to matrix-matrix products, by treating the vector as a special case of a matrix with one column.

Subsection The Matrix-Vector Product

The matrix-vector product we defined in Preview Activity 5.1 for a specific example generalizes in a very straightforward manner, and provides a convenient way to represent a system of linear equations of any size using matrices and vectors. In addition to providing us with an algebraic approach to solving systems via matrices and vectors — leading to a powerful geometric relationship between solution sets of homogeneous and non-homogeneous systems — this representation allows us to think of a linear system from a dynamic perspective, as we will see later in the section on matrix transformations.

The matrix-vector product \(A \vx\) is a linear combination of the columns of \(A\) with weights from \(\vx\text{.}\) To define this product in general, we will need a little notation. Recall that a matrix is made of rows and columns — the entries reading from left to right form the rows of the matrix and the entries reading from top to bottom form the columns. For example, the matrix

has three rows and four columns. The number of rows and columns of a matrix is called the size of the matrix, so \(A\) is a 3 by 4 matrix (also written as \(3 \times 4\)). We often need to have a way to reference the individual entries of a matrix \(A\text{,}\) and to do so we typically give a label, say \(a_{ij}\) to the entry in the \(i\)th row and \(j\)th column of \(A\text{.}\) So in our example we have \(a_{23}=7\text{.}\) We also write \(A = [a_{ij}]\) to indicate a matrix whose \(i,j\)th entry is \(a_{ij}\text{.}\) At times it is convenient to write a matrix in terms of its rows or columns. If \(A = [a_{ij}]\) is an \(m \times n\) matrix, then we will write

or, if we let \(\vr_1, \vr_2, \ldots, \vr_m\) denote the rows of the matrix \(A\text{,}\) then we can write \(A\) as ⁹

We can also write \(A\) in terms of its columns, \(\vc_1, \vc_2, \ldots, \vc_n\text{,}\) as

In general, the product of a matrix with a vector is defined as follows.

Subsection The Matrix-Vector Form of a Linear System

Subsection Homogeneous and Nonhomogeneous Systems

As we saw before, the systems with all the right hand side constants being 0 are special in that they always have a solution. (Why?) So we might consider grouping systems into two types: Those of the form \(A \vx = \vb\text{,}\) where not all of the entries of the vector \(\vb\) are \(0\text{,}\) and those of the form \(A \vx = \vzero\text{,}\) where \(\vzero\) is the vector of all zeros. Systems like \(A \vx = \vb\text{,}\) where \(\vb\) contains at least one non-zero entry, are called nonhomogeneous systems, and systems of the form \(A \vx = \vzero\) are called homogeneous systems. For every nonhomogeneous system \(A \vx = \vb\) there is a corresponding homogeneous system \(A \vx = \vzero\text{.}\) We now investigate the connection between the solutions to the nonhomogeneous system and the corresponding homogeneous system.

As we saw in the above activity, there is a relationship between solutions of a nonhomogeneous and the corresponding homogeneous system. Let us formalize this relationship. If the general solution of a system involves free variables, we can represent the solutions in parametric vector form to have a better idea about the geometric representation of the solution set. Suppose the solution is that \(x_3\) is free, \(x_2 = -2+x_3\text{,}\) and \(x_1 = 2-3x_3\text{.}\) In vector form, we can represent this general solution as

From this representation, we see that the solution set is a line through the origin (formed by multiples of \(\left[ \begin{array}{r} -3 \\ 1 \\ 1 \end{array} \right]\)) shifted by the added vector \(\left[ \begin{array}{r} 2 \\ -2 \\ 0 \end{array} \right]\text{.}\) The solution to the homogeneous system on the other does not have the shift.

Algebraically, we see that every solution to the nonhomogeneous system \(A\vx = \vb\) can be written in the form \(\vp + \vv_h\text{,}\) where \(\vp\) is a particular solution to \(A\vx = \vb\) and \(\vv_h\) is a solution to the corresponding homogeneous system \(A\vx = \vzero\text{.}\)

To understand why this always happens, we will verify the result algebraically for an arbitrary \(A\) and \(\vb\text{.}\) Assuming that \(\vp\) is a particular solution to the nonhomogeneous system \(A\vx=\vb\text{,}\) we need to show that:

• if \(\vv\) is an arbitrary solution to the nonhomogeneous system, then \(\vv = \vp + \vv_h\text{,}\) where \(\vv_h\) is some solution to the homogeneous system \(A\vx = \vzero\text{,}\) and

• if \(\vv_h\) is an arbitrary solution to the homogeneous system, then \(\vp + \vv_h\) is a solution to the nonhomogeneous system.

To verify the first condition, suppose that \(\vv\) is a solution to the nonhomogeneous system \(A \vx = \vb\text{.}\) Since we want \(\vv=\vp+\vv_h\text{,}\) we need to verify that \(\vv-\vp\) is a solution for the homogeneous system so that we can assign \(\vv_h=\vv-\vp\text{.}\) Note that

using the distributive property of matrix-vector product over vector addition. Hence \(\vv\) is of the form \(\vp + \vv_h\) with \(\vv_h = \vzero\text{.}\)

To verify the second condition, consider a vector of the form \(\vp + \vv_h\text{,}\) where \(\vv_h\) is a homogeneous solution. We have

and so \(\vp + \vv_h\) is a solution to \(A \vx = \vb\text{.}\)

Our work above proves the following theorem.

Subsection The Geometry of Solutions to the Homogeneous System

There is a simple geometric interpretation to the solution set of the homogeneous system \(A \vx = \vzero\) based on the number of free variables that imposes a geometry on the solution set of the corresponding nonhomogeneous system \(A \vx = \vb\) (when consistent) due to Theorem 5.4.

Our work in the above activity shows the geometric shape of the solution set of a consistent nonhomogeneous system is the same as the geometric shape of the solution set of the corresponding homogeneous system. The only difference between the two solution sets is that one is a shifted version of the other.

Subsection Examples

What follows are worked examples that use the concepts from this section.

Subsection Summary

• If \(A = [\vc_1 \ \vc_2 \ \cdots \ \vc_n]\) is an \(m \times n\) matrix with columns \(\vc_1\text{,}\) \(\vc_2\text{,}\) \(\ldots\text{,}\) \(\vc_n\text{,}\) and if \(\vx = \left[ \begin{array}{c} x_1\\x_2\\ \vdots \\ x_n \end{array} \right]\) is a vector in \(\R^n\text{,}\) then the matrix-vector product \(A \vx\) is defined to be the linear combination of the columns of \(A\) with corresponding weights from \(\vx\) — that is

  \begin{equation*} A \vx = x_1 \vc_1 + x_2 \vc_2 + \cdots + x_n \vc_n\text{.} \end{equation*}

• A linear system

  \begin{alignat*}{5} {a_{11}}x_1 \amp {}+{} \amp {a_{12}}x_2 \amp {}+{} \amp \cdots \amp {}+{} \amp {a_{1n}}x_n \amp {}={} \amp b_1\\ {a_{21}}x_1 \amp {}+{} \amp {a_{22}}x_2 \amp {}+{} \amp \cdots \amp {}+{} \amp {a_{2n}}x_n \amp {}={} \amp b_2\\ {} \amp {} \amp {} \amp {} \amp \vdots \ \ \amp {} \amp {} \amp {} \amp {}\\ {a_{m1}}x_1 \amp {}+{} \amp {a_{2m}}x_2 \amp {}+{} \amp \cdots \amp {}+{} \amp {a_{mn}}x_n \amp {}={} \amp b_m \end{alignat*}

  can be written in matrix form as

  \begin{equation*} A \vx = \vb\text{,} \end{equation*}

  where

  \begin{equation*} A = \left[ \begin{array}{cccc} a_{11} \amp a_{12} \amp \cdots \amp a_{1n} \\ a_{21} \amp a_{22} \amp \cdots \amp a_{2n} \\ \vdots \amp \amp \ddots \amp \vdots \\ a_{m1} \amp a_{m2} \amp \cdots \amp a_{mn} \end{array} \right], \ \ \ \vx = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right], \ \ \ \text{ and } \ \ \ \vb = \left[ \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_m \end{array} \right]\text{.} \end{equation*}

• The matrix equation \(A \vx = \vb\) has a solution if and only if \(\vb\) is a linear combination of the columns of \(A\text{.}\)

• The system \(A \vx = \vb\) is consistent for every vector \(\vb\) if every row of \(A\) contains a pivot.

• A homogeneous system is a system of the form \(A \vx = \vzero\) for some \(m \times n\) matrix \(A\text{.}\) Since the zero vector in \(\R^n\) satisfies \(A \vx = \vzero\text{,}\) a homogeneous system is always consistent.

• A homogeneous system can have one or infinitely many different solutions. The homogeneous system \(A \vx = \vzero\) has exactly one solution if and only if each column of \(A\) is a pivot column.

• The solutions to the consistent nonhomogeneous system \(A \vx = \vb\) have the form \(\vp + \vv_h\text{,}\) where \(\vp\) is a particular solution to the nonhomogeneous system \(A \vx = \vb\) and \(\vv_h\) is a solution to the homogeneous system \(A \vx = \vzero\text{.}\) In other words, the solution space to a consistent nonhomogeneous system \(A \vx = \vb\) is a translation of the solution space of the homogeneous system \(A \vx = \vzero\) by a particular solution to the nonhomogeneous system.

Finally, we argued an important theorem.

We will continue to add to this theorem, so it is a good idea for you to begin now to remember the equivalent conditions of this theorem.

Subsection Project: Input-Output Models

There are two basic types of input-output models: closed and open. The closed model assumes that all goods produced are consumed within the economy — no trading takes place with outside entities. In the open model, goods produced within the economy can be traded outside the economy.

To work with a closed model, we use an example (from Input-Output Economics by Wassily Leontief). Assume a simple three-sector economy consisting of agriculture (growing wheat), manufacturing (producing cloth), and households (supplying labor). Each sector of the economy relies on goods from the other sectors to operate (e.g., people must eat to work and need to be clothed). To model the interactions between the sectors, we consider how many units of product is needed as input from one sector to another to produce one unit of product in the second sector. For example, assume the following:

• to produce one unit (say dollars worth) of agricultural goods requires 25% of a unit of agricultural output, 28% of a unit of manufacturing output, and 27% of a unit of household output;

• to produce one unit of manufactured goods requires 20% of a unit of agricultural output, 60% of a unit of manufacturing output, and 60% of a unit of household output;

• to produce one unit of household goods requires 55% of a unit of agricultural output, 12% of a unit of manufacturing output, and 13% of a unit of household output.

These assumptions are summarized in Table 5.7.

This model is said to be closed because all good produced are used up within the economy. If there are goods that are not used within the economy the model is said to be open. Open models will be examined later.

The economist's goal is to determine what level of production in each section meets the following requirements:

• the production from each sector meets the needs of all of the sectors and

• there is no overproduction.

In general, a matrix derived from a table like Table 5.7 is called a consumption matrix, which we will denote as \(C\text{.}\) (In the example discussed here \(C = \left[\begin{array}{ccc} 0.25 \amp 0.28 \amp 0.27 \\ 0.20 \amp 0.60 \amp 0.60 \\ 0.55 \amp 0.12 \amp 0.13 \end{array} \right]\text{.}\)) A consumption matrix \(C = [c_{ij}]\text{,}\) where \(c_{ij}\) represents the proportion of the output of sector \(j\) that is consumed by sector \(i\text{,}\) satisfies two important properties.

• Since no sector can consume a negative amount or an amount that exceeds the output of another sector, we must have \(0 \leq c_{ij} \leq 1\) for all \(i\) and \(j\text{.}\)

• If there are \(n\) sectors in the economy, the fact that all output is consumed within the economy implies that \(c_{1j}+c_{2j}+ \cdots + c_{nj} = 1\text{.}\) In other words, the column sums of \(C\) are all 1.

In our example, if we let \(x = \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right]\text{,}\) then we can write the equations that guarantee that the production levels satisfy the two economists' goal in matrix form as

Now we can rephrase the question to be answered as which production vectors \(\vx\) satisfy equation (5.5). When \(C\vx = \vx\text{,}\) then the system is in equilibrium, that is output exactly meets needs. Any solution \(\vx\) that satisfies (5.5) is called a steady state solution.

So far, we considered the case where the economic system was closed. This means that the industries that were part of the system sold products only to each other. However, if we want to represent the demand from other countries, from households, capital building, etc., we need an open model. In an article in the Scientific American Leontief organized the 1958 American economy into 81 sectors. The production of each of these sectors relied on production from the all of the sectors. Here we present a small sample from Leontief's 81 sectors, using Petroleum, Textiles, Transportation, and Chemicals as our sectors of the economy. Leontief's model assumed that the production of 1 unit of output of

• petroleum requires 0.1 unit of petroleum, 0.2 units of transportation, and 0.4 units of chemicals;

• textiles requires 0.4 units of petroleum, 0.1 unit of textiles, 0.15 units of transportation, and 0.3 units of chemicals;

• transportation requires 0.6 units of petroleum, 0.1 unit of transportation, and 0.25 units of chemicals;

• chemicals requires 0.2 units of petroleum, 0.1 unit of textiles, 0.3 units of transportation, and 0.2 units of chemicals.

A summary of this information is in Table 5.8. Assume the units are measured in dollars.

In the open model, there is another part of the economy, called the open sector, that does not produce goods or services but only consumes them. If this sector (think end consumers, for example) demands/consumes \(d_1\) units of Petroleum, \(d_2\) units of Textiles, \(d_3\) units of Transportation, and \(d_4\) units of Chemicals, we put this into a final demand vector \(\vd = \left[ \begin{array}{c} d_1 \\d_2 \\ d_3 \\ d_4 \end{array} \right]\text{.}\)

An economist would want to find the production level where the demand from the good/service producing sectors of the economy plus the final demand from the open sector exactly matches the output in each of the sectors. Let \(x_1\) represent the number of units of petroleum output, \(x_2\) the number of units of textiles output, \(x_3\) the number of units of transportation output, and \(x_4\) the number of units of chemical output during any time period. Then the production vector is \(\vx = \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\x_4 \end{array} \right]\text{.}\) So an economist wants to find the production vectors \(\vx\) such that

where \(\vd = \left[ \begin{array}{c} d_1 \\d_2 \\ d_3 \\ d_4 \end{array} \right]\) is the demand vector from the open market. The matrix

derived from Table 5.8, is called the exchange matrix.