Month Semiprivate Semiprivate

9/27/2018 Module 2

https://onlinecampus.bu.edu/bbcswebdav/pid-6253499-dt-content-rid-23201016_1/courses/18fallmetad510_d1/course/module2/allpages.htm 1/47

Module 2

This is a single, concatenated file, suitable for printing or saving as a PDF for offline viewing. Please

note that some animations or images may not work.

Module 2 Study Guide and Deliverables

Topics and

Readings

Lecture 3 (Sept 20): Systems of Linear Equations and Matrices

Module 2 Lecture 3 lecture notes

Textbook, Chapter 6

Lecture 4 (Sept 27): Linear Programming and Its Applications

Module 2 Lecture 4 lecture notes

Textbook, Chapter 7 (skip 7.4, 7.6, and 7.7)

Discussions: Module 2 Discussion postings due Oct 3 at 11:59 PM ET

Assignments: Complete the Individual Exercise problems at the end of each Lecture, as well as

selected problems from the textbook.

Midterm Midterm 1 (Oct 4) Covering topics from Lectures 1 through 4

Lecture 3 – Systems of Linear Equations and Matrices

Learning Objectives

After successfully completing the module, students will be able to:

1. Solve systems of linear equations

2. Apply augmented matrices and the Gauss Jordan Method

3. Calculate Matrix operations and Inverses

4. Calculate production matrices from demand and input-output matrices

Systems of Linear Equations

A system of linear equations is simply a group of linear equations on some number of variables, typically denoted . By linear we mean

that the only operations permitted in the system are addition, subtraction, and multiplication by constants. For example,

qualifies as a system of linear equations. On the other hand, none of the following equations would be found in a system of linear equations:

x, y, z

4x + y = z

y = 3x

x + y + z = 12

9/27/2018 Module 2

https://onlinecampus.bu.edu/bbcswebdav/pid-6253499-dt-content-rid-23201016_1/courses/18fallmetad510_d1/course/module2/allpages.htm 2/47

Given a system of two linear equations in two variables, the most common ways of solving it are via substitution and elimination.

Substitution

Suppose you’re given

To solve this via the substitution method, isolate one variable, then substitute the new equation from the isolated variable into the other

equation, as follows:

Plugging this into the first equation, we get

We can now solve for . Simplifying:

We can then plug 4.5 in for in either original equation to get

Elimination

The elimination method involves multiplying both sides of an equation by a constant, then adding the resulting equations to cancel a variable.

For example, given the following system of equations:

We could first multiply the bottom equation through by -3 (this would help us to eliminate x from the system), resulting in:

Summing the two equations yields:

Which gives us . Substituting 5 for in either of the original equations returns .

4xy = z

y = + 12×2

y = ex

3x − 4y = 15

2x + 5 = 6y

2x + 5 = 6y ⟹ x = 6y − 5

2

3 × − 4y = 15 6y − 5

2

y

9y − 7.5 − 4y = 15 ⟹ 5y = 22.5 ⟹ y = 4.5

y

2x + 5 = 6 × 4.5 ⟹ x = (27 − 5)/2 ⟹ x = 11

3x + 4y = 26

x − 7y = −33

3x + 4y = 26

−3x + 21y = 99

0x + 25y = 125

y = 5 y x = 2

9/27/2018 Module 2

https://onlinecampus.bu.edu/bbcswebdav/pid-6253499-dt-content-rid-23201016_1/courses/18fallmetad510_d1/course/module2/allpages.htm 3/47

Uniqueness, Dependency, and Consistency

If, like the previous two examples, substitution and elimination, a system has a unique solution it is said to be independent.

Systems of equations don’t always have unique answers. Take the following system of equations:

Dividing the bottom equation through by 3 and bringing over to the other side yields:

You can see that any value of can result in an answer. For example, ( , ) and ( , ) both work. Such a system is

called dependent.

A system is said to be inconsistent if there is no solution that satisfies all of its equations.

Take the following system of linear equations as an example:

Dividing the bottom equation by 4, we get:

This means that has to equal both 40 and 25: an impossibility.

Figure 1 shows, from left to right, graphs of dependent, inconsistent and independent systems of 2 equations.

Figure 1: Dependent, Inconsistent and Independent Systems of 2 Equations

Larger Systems of Linear Equations

x − 3y = 10

3x − 30 = 9y

y

x − 3y = 10

x − 3y = 10

x x = 16 y = 2 x = 10 y = 0

5x + 6y = 40

20x + 24y = 100

5x + 6y = 25.

5x + 6y

9/27/2018 Module 2

https://onlinecampus.bu.edu/bbcswebdav/pid-6253499-dt-content-rid-23201016_1/courses/18fallmetad510_d1/course/module2/allpages.htm 4/47

Elementary Operations

A system of linear equations can contain more than two equations. To solve such an equation, we rely on the elementary operations. These

operations, used correctly, make a system easier to solve without changing the solution. If two linear equations have the same solution, they’re

called equivalent.

The elementary operations are:

1. Change the order of equations

Eq. A

Eq. B

Eq. C

 

 

Eq. A’ = B

Eq. B’ = A

Eq. C’ = C

2. Multiply an equation by a nonzero constant

Eq. A

Eq. B

Eq. C