7.3 Gaussian Elimination with Back-Substitution Gaussian Elimination is the FIRST of FIVE methods we will learn about for solving systems of equations. The main goal in using the Gaussian Elimination method is to transform your system of equations into row-echelon form (stair-step pattern). This is done by performing different Elementary Row Operations on the system of equations. Example 1: Solve the following system using the elementary row operations

Finding the solution: Back solution

X =

Y =

Z =

Row Operations in Gaussian Elimination

1. Replace any row with a linear combination of that row and any other row. (

R2 2R1 R2)

2. Interchange any two rows. (

R1R2)

3. Replace any row with a nonzero multiple of that row.

(

2R2 R2)

Matrix = a rectangular array of elements.

232221

131211

32854

302

aaa

aaaA

rows

columns

element a21

(row 2, column 1)

dimensions

(2 rows by 3 columns)

Augmented Matrices (a way to represent a linear system)

332

54

yx

yx

332

541

Optional Scratch Work

Example 2. Using Matrices to Represent a System of Linear Equations Write the augmented Matrix that represents each system and give the dimensions (order) of the

matrix.

(a) 74

932

b

ba (b)

x 2y 3z 9

x 3y 4

2x 5y 5z 17

Example 3: Solve the following system using Gaussian Elimination

Optional Scratch Work System of linear Equations

Matrix Form Optional Scratch Work

Turn the system into an augmented matrix

Back substitution to find the solution

Possible outcomes for solving a system of equations:

Exactly One of the Below is true: Example

1.

1

2

343

z

zy

zyx

OR

If the system of equations can be reduced to row echelon form (stair-step pattern), there is exactly one solution. Use back substitution to solve for x, y, z. Solution: (-4, 1, 1).

2.

20

2

343

zy

zyx

OR

If the system of equations reduces to an equality that is not true, there is no solution.

3.

00

2

343

zy

zyx

OR

If the system of equations reduces to more variables than there are variables (e.g. 2 equations, 3 variables above), there are infinitely many solutions. If so, set z=a, where a is any real number and use back substitution as if there were exactly one solution. Final answers will have x, y, z in terms of a. Solution: (-3-a, 2-a, a) For all real numbers, a.

Homework: Worksheet 7.3