Sections 1.1 & 1.2 Intro to Systems of Linear Equations & Gauss Jordan Elimination

Sections 1.1 & 1.2 Intro to Systems of Linear Equations

& Gauss Jordan Elimination

Linear Algebra – Linear Equationsy = 3x – 2 2x – 3y + 4z = 9

Vector Spaces

A solution to an equation.

A solution to multiple equations (a system of equations).

A parametric representation of a solution.

Geometrical meaning of a solution.

For two equations in two unknowns:One Solution No Solutions Infinite Solutions

2x + 3y = 6 2x + 3y = 6 2x + 3y = 6 4x – 2y = 12 4x + 6y = –24 4x + 6y = 12


For three equations in three unknowns:


For three equations in three unknowns:

The number of solutions to a system of linear equations.

A system is consistent if there is at least one solution to it.

A system is inconsistent if there is no solution for it.

Equivalent systems – systems that have the same solution set.

Hard to solve: Easy to solve:

x = 1 y = –1 z = 2

x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17

+ 0y + 0z0x + + 0z0x + 0y +

Hard to solve: Easy to solve:

x = 1 y = –1 z = 2

x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17

12 3

Elementary Operations:1. Interchange any two equations.2. 3.

x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17

–2x + 3y + 7z = –4 x – 2y + 3z = 9 2x – 5y + 5z = 17

Elementary Operations:1. 2. Multiply any equation by a nonzero scalar.3.

x – 2y + 3z = 9–2x + 3y + 7z = –420x – 50y + 50z = 170

x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17

Elementary Operations:1. 2. 3. Add a multiple of one equation to another equation.

x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17

x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17

2x – 4y +12z = 18–2x + 3y + 7z = –4 2x – 5y + 5z = 17

2x – 4y + 12z = 18 – y + 19z = 14 2x – 5y + 5z = 17 x – 2y + 3z = 9 – y + 19z = 14 2x – 5y + 5z = 17

x = 1 y = –1 z = 2

x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17

+ 0y + 0z0x + + 0z0x + 0y +

Gauss Jordan Elimination:

Gaussian Elimination:

x – 2y + 3z = 9 y + 3z = 5 z = 2

x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17

0x + 0x + 0y +

Ex. Use Gauss Jordan elimination to solve the system x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17

x – 2y + 3z = 9 y + 3z = 5 2x – 5y + 5z = 17

x – 2y + 3z = 9 y + 3z = 5 –y – z = –1

x – 2y + 3z = 9 y + 3z = 5 2z = 4

x + 9z = 19 y + 3z = 5 2z = 4

(method of elimination)


x + 9z = 19 y + 3z = 5 2z = 4

x + 9z = 19 y + 3z = 5 2z = 4

x + 9z = 19 y + 3z = 5 z = 2

x + 9z = 19 y = –1 z = 2

x = 1 y = –1 z = 2


Notice that we can use elementary row operations on a matrix to work through the Gauss Jordan method of elimination. x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17

x – 2y + 3z = 9 y + 3z = 52x – 5y + 5z = 17

x – 2y + 3z = 9 y + 3z = 5 –y – z = –1

x – 2y + 3z = 9 y + 3z = 5 2z = 4

x + 9z = 19 y + 3z = 5 2z = 4

1 2 3 9

0 1 3 5

2 5 5 17

1 2 3 9

0 1 3 5

0 1 1 1

1 2 3 9

0 1 3 5

0 0 2 4

1 0 9 19

0 1 3 5

0 0 2 4

1 2 3 9

1 3 0 4

2 5 5 17

R2+R1→R2

R3–2R1→R3

R3+R2→R3

R1+2R2→R3

x + 9z = 19 y + 3z = 5 2z = 4

1 0 9 19

0 1 3 5

0 0 2 4


x + 9z = 19 y + 3z = 5 2z = 4

x + 9z = 19 y + 3z = 5 z = 2

x = 1 y + 3z = 5 z = 2

x = 1 y = –1 z = 2

1 0 9 19

0 1 3 5

0 0 2 4

1 0 9 19

0 1 3 5

0 0 1 2

1 0 0 1

0 1 3 5

0 0 1 2

1 0 0 1

0 1 0 1

0 0 1 2


½ R3→R3

R1–9R3→R1

R2–3R3→R2

Elementary Row Operations(We are allowed to use these operations on a matrix when trying to solve a system of linear equations.)

Elementary Row Operation: Notation:1. 2. 3.

1. Interchange two rows Ri ↔ Rk

2. Multiply (or divide) a row by a nonzero constant. cRi → Ri

3. Add (or subtract) a multiple of one row to another. Ri + cRk → Ri

We'd like to take our original augmented matrix, and through row operations put it in the following form:

where the bi's are just some constants (some numbers).

1

2

3

4

1

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1n

n

b

b

b

b

b

b

1 2 3 7

0 0 0 0

0 0 0 0

1 0 5 2

0 1 6 3

For example, this matrix has a solution that is easy to see, (1, 3, 5), because the matrix is in the final form that we want.

1 0 0 1

0 1 0 3

0 0 1 5

This is not always possible though. The following are matrices that cannot be put into this form.

Reduced Row Echelon FormA matrix is said to be in reduced echelon form if all of the following properties hold true:1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.

Reduced Row Echelon FormA matrix is said to be in reduced echelon form if all of the following properties hold true:1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.

x = 1 y = –1 z = 2

x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17

Gauss Jordan Elimination:

Gaussian Elimination:

x – 2y + 3z = 9 y + 3z = 5 z = 2

x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17

1 2 3 9

1 3 0 4

2 5 5 17

1 2 3 9

0 1 3 5

0 0 1 2

1 0 0 1

0 1 0 1

0 0 1 2

Reduced Row Echelon Form

Row Echelon Form

Ex. Determine which of the following matrices are in reduced row echelon form.

(a) (b) (c) 1 0 0 4

0 1 0 5

0 0 1 2

1 0 0 3

0 1 0 4

0 0 0 0

0 0 1 5

1 0 0 2

0 0 1 4

0 1 0 1

Reduced Row Echelon Form1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.


(d) (e) (f) 1 0 0 0 4

0 0 1 0 5

0 0 0 1 7

1 0 3 4

0 1 2 1

0 0 0 0

1 0 3

0 1 2

0 0 0



(g) (h)1 0 0 2

0 1 0 4

2 0 0 3

0 1 0 5

0 0 1 7


Ex. Put the matrix into reduced row echelon form.

1 3 11

3 4 6

2 7 17

A

1 3 11

3 4 6

2 7 17

1 3 11

0 13 39

0 13 39

1 3 11

0 1 3

0 13 39

R2 – 3R1→R2

R3 – 2R1→R3

R1 – 3R2→R1

R3 + 13R2→R3

1 0 2

0 1 3

0 0 0

–1/13 R2→R2

Using a calculator with matrices.

Ex. Solve the following system. x + y = 11 3x – 4y = –6 2x – 7y = –17

1 1 11

3 4 6

2 7 17

1 1 11

0 7 39

0 9 39

38

7

397

787

1 0

0 1

0 0

1 0 0

0 1 0

0 0 1

Ex. Solve the following system of equations. x1 – 4x2 + 7x3 = –3

–2x1 + 9x2 – 4x3 = 7

x1 – 3x2 + 17x3 = –21 4 7 3

2 9 4 7

1 3 17 2

1 4 7 3

0 1 10 1

0 1 10 1

1 0 47 1

0 1 10 1

0 0 0 0

Here's one view of the three planes: Here's a side view of the three planes:

Ex. Solve the following system of equations. x1 – 4x2 + 7x3 = –3

(A graph of this system is given below.) –2x1 + 9x2 – 4x3 = 7

x1 – 3x2 + 17x3 = –2

Ex. Solve the following system of equations. x1 – 2x2 + 3x3 = 4

–2x1 + 4x2 – 6x3 = –8

3x1 – 6x2 + 9x3 = 121 2 3 4

2 4 6 8

3 6 9 12

1 2 3 4

0 0 0 0

0 0 0 0

Here's one view of the three planes: Here's a side view of the three planes:

Ex. Solve the following system of equations. x1 – 2x2 + 3x3 = 4

(A graph of this system is given below.) –2x1 + 4x2 – 6x3 = –8

3x1 – 6x2 + 9x3 = 12

Notation and terminology

If we call a matrix A then we shall refer to the entries in A as follows:ai j is the number in matrix A in row i column j.

If we call a matrix A then we shall refer to the entries in A as follows:ai j is the number in matrix A in row i column j.

If then we have the following:a1 2 =

a3 1 =

If then we have the following:b1 2 =

b2 3 =

11 13 3

6 5 1

4 0 2

A

1 3 4

2 2 1B

13

–4

3

1

Given a system like –5x + 6y – 2z = 11 we will often look at the following x + 3y = 2 two matrices. 2x – y + z = 4

Coefficient Matrix Augmented Matrix

5 6 2

1 3 0

2 1 1

5 6 2 11

1 3 0 2

2 1 1 4

Homogenous systems

8x – 2y + 6z = 0 9x + 3y + 7z = 0 4x – 5y + 2z = 0

13x – 8y + 2z = 0 2x + 4y –10z = 0 5x – 7y + 3z = 0 x + 3y + 5z = 0

Documents

Sections 1.1 & 1.2 Intro to Systems of Linear Equations & Gauss Jordan Elimination