We solve the following problem:

• We wish to set up a lab with computers, terminals and workstations. We require 5 terminals for every computer and a workstation for every two computers. How many of each will we be able to purchase with $280,000 if each computer costs $2000, each terminal costs $500, and each workstation costs $5000?

We begin by using the symbols x₁, x₂ and x₃ to stand for the respective numbers of computers, terminals and workstations.

We next symbolize the total costs of the computers, of the terminals and of the workstations.

We finally write an equation, using these symbols, for the total cost with the given total cost.

At this point we have one equation and three unknowns. We need two more independent equations relating our three unknowns.

• We use the conditions relating the numbers of terminals and workstations to the numbers of computers to obtain two more equations.

• We see that the number x₂ of terminals is five times the number x₁ of computers, so x₂ = 5 x₁.
• The number x₃ of workstations is double the number x₁ of computers, so x₃ = 2 x₁.

• In order to align our three equations, we will express these two new equations, respectively, as -5 x₁ + x₂ = 0 and -2 x₁ + x₃ = 0.

• We write our system as in the figure below.

We now consider the following problem:

• We are to create a daily cafeteria diet consisting of gelatin, fish and meet. Each unit of larded gelatin contains 10 grams of carbohydrates, 1 gram of protein and 50 grams of fat, each unit of fish food contains 50 grams of carbohydrates, 3 grams of protein and 2 grams of fat, and each unit of artificial meat contains 200 grams of carbohydrates, .2 grams of protein and no fat. Our meal is to consist of 600 grams of carbohydrates, 20 grams of protein and 200 grams of fat.

We begin by letting x₁, x₂ and x₃ stand for the number of grams of gelatin, fish and meat in the daily diet.  We wish to write three equations using these three unknowns.

• We write an equation in which we set a symbolic expression for the number of grams of carbohydrates equal to the desired 600.
• There are 10 x₁ grams of carbohydrate in the x₁ grams of gelatin, 50 x₂ grams of carbohydrate in the x₂ grams of fish, and 200 x₃ grams carbohydrate in the x₃ grams of meat.
• The symbolic expression for the number grams carbohydrate is therefore 10 x₁ + 50 x₂ + 200 x₃.
• We set this equal to 600 to obtain the first equation.

• The second equation sets the symbolic expression x₁ + 3 x₂ + .2 x₃ for the number of grams of protein equal to the desired 20.

• The third equation sets the symbolic expression 50 x₁ + 10 x₂ for the number of grams of fat equal to the desired 200.

Video File #01

We now a matrix representation to depict this system of equations.

• We first write the system as a product of a matrix A and a column vector x, with x representing the number of grams of gelatin, fish and meat respectively and A representing the amounts of carbohydrate, protein and fat in each unit of each component.
• To complete the system we set this matrix product equal to the column vector B representing the desired amounts of carbohydrate, protein and fat.

The system is depicted below.

• We can write the system as A x = B, with the matrices A and B as indicated.
• We could solve the system if we could find the matrix A^-1.

In the figure below we represent the rules H⁺ = 1.2 H - .3 W and W⁺ = .2 H + .7 W representing the numbers of hares and wolves in an ecosystem.

• The 1.2 H indicates that each living hare will, by reproduction, contribute 1.2 hares to the subsequent population during a transition period.
• The -.3 W indicates that each wolf will eat .3 hares during a transition period.
• The .7 W indicates that each living wolf will, by reproduction, contribute .7 wolves to the subsequent population during a transition period.
• The .2 H indicates that each living hare will supply the nourishment to allow .2 wolves to survive for a transition period.
• These numbers are not particularly realistic, but they serve to illustrate the model.

We express these rules in matrix form as shown.

• We wish to find the coefficient matrix such that when we multiply the column vector [ H W ]`representing the present number of hares and wolves by the matrix, we obtain the numbers after a transition period.
• We see that the coefficients must be as in the bottom line below.

Video File #02

The matrix equation below represents the rules

• p₁⁺ = 1/4 p₁ + 1/2 p₂ + 1/4 p₃, indicating the probability p₁⁺ that the stock market will be up tomorrow in terms of the probabilities p₁ that the market is up today, p₂ that the market is down today, and p₃ that the market is unchanged today,
• p₂⁺ = 1/2 p₁ + 1/4 p₂ + 1/2 p₃, indicating the probability p₂⁺ that the market will be down tomorrow in terms of the same p₁, p₂ and p₃, and
• p₃⁺ = 1/4 p₁ + 1/4 p₂ + 1/4 p₃, indicating in a similar manner the probability p₃⁺ that the market will be unchanged tomorrow.

The figure below illustrates the probability p₁⁺ = P(up tomorrow).

• According to the rules the probability that the market is up tomorrow will be 1/4 the probability that it is up today, plus 1/2 the probability that it is down today plus 1/4 the probability that it is the same today.

We note that the transition matrix is what we call a 'stochastic' matrix, where the columns of the matrix add up to 1.

Video File #03

We define the dot product between two vectors as in the figure below as the product | a | | b | with the cosine of the angle between the vectors.

• If we represent the first vector a as a row vector and the second vector b as a column vector, it turns out that the dot product is equal to the vector product a  ^.  b.

In the figure below we calculate this dot product, as well as the magnitudes | a | and| b | of the two vectors.

In the figure below we rearrange the definition of the dot product to find the angle between the vectors in terms of the calculations of the preceding figure.