Download Chapter 1 Notes

Sections 1.1 & 1.2
Intro to Systems of Linear Equations
& Gauss Jordan Elimination
Linear Algebra – Linear Equations
y = 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
2x + 3y = 6
4x – 2y = 12
2x + 3y = 6
4x + 6y = –24
Infinite Solutions
2x + 3y = 6
4x + 6y = 12
Geometrical meaning of a solution.
For three equations in three unknowns:
Geometrical meaning of a solution.
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:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
Easy to solve:
→
x + 0y + 0z = 1
0x + y + 0z = –1
0x + 0y + z = 2
Hard to solve:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
1
2
3
Easy to solve:
x
→
y
=1
= –1
z =2
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 = –4
2x – 5y + 5z = 17
x – 2y + 3z = 9
–2x + 3y + 7z = –4
20x – 50y + 50z = 170
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
Gauss Jordan Elimination:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
→
x + 0y + 0z = 1
0x + y + 0z = –1
0x + 0y + z = 2
Gaussian Elimination:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
→
x – 2y + 3z = 9
0x + y + 3z = 5
0x + 0y + z = 2
Ex. Use Gauss Jordan elimination to solve the system
(method of elimination)
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
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
Ex. Use Gauss Jordan elimination to solve the system
x
+ 9z = 19
y + 3z = 5
2z = 4
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
Ex. Use Gauss Jordan elimination to solve the system
x
+ 9z = 19
y + 3z = 5
2z = 4
x+
9z = 19
y + 3z = 5
z=2
x+
y
9z = 19
= –1
z=2
y
=1
= –1
z =2
x
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
Notice that we can use elementary row operations on a matrix to work through the
Gauss Jordan method of elimination.
 1 2 3 9 
x – 2y + 3z = 9
 1 3 0 4 
–x + 3y
= –4


2x – 5y + 5z = 17
 2 5 5 17 
x – 2y + 3z = 9
1 2 3 9 
y + 3z = 5
R2+R1→R2  0 1 3 5 


2x – 5y + 5z = 17
 2 5 5 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
R3–2R1→R3
1 2 3 9 
0 1 3 5 


0 1 1 1
R3+R2→R3
1 2 3 9 
0 1 3 5 


0 0 2 4 
1 0 9 19 
0 1 3 5 


R1+2R2→R3 0 0 2 4 
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
+ 9z = 19
y + 3z = 5
2z = 4
1 0 9 19 
0 1 3 5 


0 0 2 4 
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
+ 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
z=2
1 0 9 19 
0 1 3 5 


½ R3→R3 0 0 1 2 
x
=1
y + 3z = 5
z=2
x
y
=1
= –1
z =2
R1–9R3→R1 1 0 0 1 
0 1 3 5 


 0 0 1 2 
R2–3R3→R2
1 0 0 1 
 0 1 0 1


 0 0 1 2 
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:
1. Interchange two rows
2. Multiply (or divide) a row by a nonzero constant.
3. Add (or subtract) a multiple of one row to another.
Notation:
Ri ↔ Rk
cRi → Ri
Ri + cRk → Ri
We'd like to take our original augmented matrix, and through row operations
put it in the following form:
1 0 0 0
0 0 b1 


b
0
1
0
0
0
0
2 

0 0 1 0
0 0 b3 


b
0
0
0
1
0
0
4 







0 0 0 0
1 0 bn1 


b
0
0
0
0
0
1

n 

where the bi's are just some constants (some numbers).
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.
1 2 3 7 


0
0
0
0


0 0 0 0 
1 0 5 2 


0
1
6
3


Reduced Row Echelon Form
A 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 Form
A 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.
Gauss Jordan Elimination:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
x
→
y
=1
= –1
z =2
Gaussian Elimination:
x – 2y + 3z = 9
–x + 3y
= –4
2x – 5y + 5z = 17
 1 2 3 9 
 1 3 0 4 


 2 5 5 17 
→
x – 2y + 3z = 9
y + 3z = 5
z =2
1 0 0 1 
 0 1 0 1


 0 0 1 2 
Reduced Row Echelon Form
1 2 3 9 
0 1 3 5 


0 0 1 2 
Row Echelon Form
Ex. Determine which of the following matrices are in reduced row echelon form.
(a) 1 0 0 4 
(b) 1 0 0 3 
(c) 1 0 0 2 






0
1
0
5
0 1 0 4 
0
0
1
4






0
0
0
0
0 0 1 2 
0 1 0 1 


0 0 1 5 
Reduced Row Echelon Form
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.
Ex. Determine which of the following matrices are in reduced row echelon form.
(d) 1 0 0 0 4 


0
0
1
0
5


0 0 0 1 7 
(e)
1 0 3 4 


0
1
2
1


0 0 0 0 
(f)
1 0 3 


0
1
2


0 0 0 
Reduced Row Echelon Form
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.
Ex. Determine which of the following matrices are in reduced row echelon form.
(g)
1 0 0 2 


0
1
0
4


(h)  2 0 0 3 


0
1
0
5


0 0 1 7 
Reduced Row Echelon Form
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.
1 3 11 
Ex. Put the matrix A   3 4 6  into reduced row echelon form.


 2 7 17 
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 
1 0

0 1
0 0
2

3
0 
R2 – 3R1→R2
R3 – 2R1→R3
–1/ R →R
13 2
2
R1 – 3R2→R1
R3 + 13R2→R3
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 
1 0
0 1

 0 0

39 
7

78 
7
38
7
1 0 0 
0 1 0 


0 0 1 
Ex. Solve the following system of equations.
 1 4 7

 2 9 4
 1 3 17
3

7
2 
1 4 7

0 1 10
0 1 10
3

1
1 
1 0 47

0 1 10
0 0 0
1

1
0 
x1 – 4x2 + 7x3 = –3
–2x1 + 9x2 – 4x3 = 7
x1 – 3x2 + 17x3 = –2
Ex. Solve the following system of equations.
(A graph of this system is given below.)
Here's one view of the three planes:
x1 – 4x2 + 7x3 = –3
–2x1 + 9x2 – 4x3 = 7
x1 – 3x2 + 17x3 = –2
Here's a side view of the three planes:
Ex. Solve the following system of equations.
 1 2 3

 2 4 6
 3 6 9
1 2 3

0 0 0
0 0 0
4

8
12 
4

0
0 
x1 – 2x2 + 3x3 = 4
–2x1 + 4x2 – 6x3 = –8
3x1 – 6x2 + 9x3 = 12
Ex. Solve the following system of equations.
(A graph of this system is given below.)
Here's one view of the three planes:
x1 – 2x2 + 3x3 = 4
–2x1 + 4x2 – 6x3 = –8
3x1 – 6x2 + 9x3 = 12
Here's a side view of the three planes:
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.
 11 13 3
If A   6 5 1  then we have the following:


 4 0 2 
a1 2 = 13
a3 1 = –4
1 3 4
If B  
then we have the following:

 2 2 1 
b1 2 = 3
b2 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
 5 6 2 
1 3 0


 2 1 1 
Augmented Matrix
 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