The document contains announcements from a class instructor. It notifies students that if they have not been able to access the class website or did not receive an email, to contact the instructor. It also reminds students that homeworks are posted on the class website and to check for any updates.

1. Announcements
If anyone has not been able to access the class website please
email me at pgeorge3@mtu.edu
If anyone didn't get the email sent via the class listserv, please
see me.
Homeworks are posted on the class website. I have made a few
changes, so please make sure you have the updated list.

2. From Yesterday
Remember the 3 equations with 3 variables from yesterday?
x − 2y + z = 0
3y − 12z = 12
−8x + 10y + 18z = −18
We want a solution (values of x , y and z ) that will satisfy all 3
equations.
We will use matrix notation that we saw yesterday
Start with the augmented matrix

3. Augmented matrix for the above system
Augmented column in circles
 

 1 −2 1 0 

 
 



0 3 −12 12 


 
−8 10 18 −18
 
 

4. Augmented matrix for the above system
Augmented column in circles
 

 1 −2 1 0 

 
 



0 3 −12 12 


 
−8 10 18 −18
 
 
Then what?

5. Augmented matrix for the above system
We want the red numbers to be zero (as many as possible)
 

 1 −2 1 0 

 
 



0 3 −12 12 


 
−8 10 18 −18
 
 

6. Augmented matrix for the above system
We want the red numbers to be zero (as many as possible)
 

 1 −2 1 0 

 
 



0 3 −12 12 


 
−8 10 18 −18
 
 
O diagonal elements in the coecient matrix should become zero

7. To achieve this
We do one or more of the 3 operations we discussed yesterday

8. To achieve this
We do one or more of the 3 operations we discussed yesterday
Add a row to a multiple of another row to replace that row
(Choose the multiple so that it gives a zero somewhere)

9. To achieve this
We do one or more of the 3 operations we discussed yesterday
Add a row to a multiple of another row to replace that row
(Choose the multiple so that it gives a zero somewhere)
Multiply a particular row by a non-zero number

10. To achieve this
We do one or more of the 3 operations we discussed yesterday
Add a row to a multiple of another row to replace that row
(Choose the multiple so that it gives a zero somewhere)
Multiply a particular row by a non-zero number
Interchange any two rows.(Useful if the diagonal becomes zero
somehow)

11. To achieve this
We do one or more of the 3 operations we discussed yesterday
Add a row to a multiple of another row to replace that row
(Choose the multiple so that it gives a zero somewhere)
Multiply a particular row by a non-zero number
Interchange any two rows.(Useful if the diagonal becomes zero
somehow)
Copy the problem correctly. Do calculations carefully. Even
small mistakes will lead you into wasting hours with no
solution in sight.

12. Ok. Let's do the row operations
Add 8 times rst row to third row to get the new third row)
 

 1 −2 1 0 

8R1+R3
 
 



0 3 −12 12 


 
−8 10 18 −18
 
 

13. Ok. Let's do the row operations
Add 8 times rst row to third row to get the new third row)
 

 1 −2 1 0 

8R1+R3
 
 



0 3 −12 12 


 
−8 10 18 −18
 
 
O diagonal elements in the coecient matrix should become zero

14. Result
Try not to lose existing zeros
 

 1 −2 1 0 

 
 



0 3 −12 12 


 
0 −6 26 −18
 
 

15. Moving on
Divide second row by 3, makes life easier
 

 1 −2 1 0 

 
 



0 3 −12 12 


 
0 −6 26 −18
 
 

16. This gives
Getting ones on the diagonal is good
 

 1 −2 1 0 

 
 



0 1 −4 4 


 
0 −6 26 −18
 
 

17. Next
Add 6 times Row 2 to Row 3 to get new Row 3
 

 1 −2 1 0 

 
 
0 1 −4 4
6R2+R3
 
 
 
 
0 −6 26 −18
 
 

18. Result
More zeros :-)
 

 1 −2 1 0 

 
 
0 1 −4 4
 
 
 
 
0 0 2 6
 
 

19. Next
Divide the last row by 2
 

 1 −2 1 0 

 
 
0 1 −4 4
 
 
 
 
0 0 1 3
 
 

20. Next
Add Row 2 to 4 times Row 3 to get new Row 2
 

 1 −2 1 0 

 
 
0 1 0 16
R2+4R3
 
 
 
 
0 0 1 3
 
 

21. Next
Add Row 1 to 2 times Row 2 to get new Row 1
 
1 −2 1 0
R1+2R2
 
 
 
 
0 1 0 16
 
 
 
 
0 0 1 3
 
 

22. Next
Almost there
 

 1 0 1 32 

 
 
0 1 0 16
 
 
 
 
0 0 1 3
 
 

23. Next
Do Row 1 - Row 3 to get new Row 1
 

 1 0 1 32 

 
R1-R3
 
0 1 0 16
 
 
 
 
0 0 1 3
 
 

24. Next
Bingo!!
 

 1 0 0 29 

 
 
0 1 0 16
 
 
 
 
0 0 1 3
 
 
The solution is staring right at us
The coecient matrix is now a TRIANGULAR matrix

25. Any preferences? Order?
None! You could do any of these operations in any order you
want
You could combine two or more operations in one step as you
practice more problems.
If your matrix is getting worse with each step, make sure you
copied the right problem and check your calculations.

26. Another example
x +y +z = 3
2x − y − z = 5
2x + 2y + 2z = 7
Vertical line separates augmented column
 

 1 1 1 3; 

 
 



2 −1 −1 5; 


 
2 2 2 7;
 
 

27. Row Operation
R3 - 2R1 to give new Row 3
 

 1 1 1 3 

 
R1-R3
 
2 −1 −1 5
 
 
 
 
0 0 0 1
 
 

28. Row Operation
R3 - 2R1 to give new Row 3
 

 1 1 1 3 

 
R1-R3
 
2 −1 −1 5
 
 
 
 
0 0 0 1
 
 
We have a problem, 0=1!!!!!

29. Inconsistent System
The above example is an inconsistent system. In other words
whenever your row reduced matrix looks like (could happen in any
row)
Here * is a non-zero number
 

 a b c d 

 
 
0 f g h
 
 
 
 
 
0 0 0
 
 ∗ 

30. Inconsistent System
If all elements in a row left to the augmented column are zero
with a non-zero element in the augmented column, the system
is inconsistent (no solutions, parallel planes)
Usually happens when an equation is multiplied by a certain
number but the right hand side is done wrong (not always)
Simple example of an inconsistent 2 equation, 2 variable system is
x +y = 1
2x + 2y = 4

31. Problem 12 sec 1.1
x − 3y + 4z = −4
3x − 7y + 7z = −8
−4x + 6y − z = 7
Vertical line separates augmented column
 

 1 −3 4 −4 

 
 
3 −7 7 −8
 
 
 
 
−4 6 −1 7
 
 

32. Problem 12 sec 1.1
R2-3R1 for new R2 and R3+4R1 for new R3
 

 1 −3 4 −4 

 
 
0 2 −5 4
 
 
 
 
0 −6 15 −9
 
 

33. Problem 12 sec 1.1
1
Do 3 R3 and add to R2 to get new R3
 

 1 −3 4 −4 

 
 
0 2 −5 4
 
 
 
 
0 0 0 1
 
 
Inconsistent

34. Problem 20 sec 1.1
Determine the value of h so that the following is the augmented
matrix of a consistent 

linear system.

 2 −3 h 

 
 
−6 9 5
 
Solution: Add 3R1 to R2 to get new R2 (Don't forget that the
augmented matrix is given to you)
 

 2 −3 h 

 
 
 

 0 0 3h + 5 

If this has to be consistent, 3h + 5 = 0 or h = −5 .
3

35. Sec 1.2, Row reduction, Echelon forms
To develop an ecient algorithm for any matrix irrespective of
whether it represents a linear system.
Nonzero row/column means a certain row/column has atleast
one nonzero entry
Leading entry of a row means the rst nonzero entry in a row
(left most)

36. Denition
Echelon form (Row Echelon form, REF): A rectangular matrix is of
Echelon form (Row Echelon Form or REF) if
All nonzero rows are ABOVE any rows with all zeros
Each leading entry of a row in a column is to the RIGHT to
the leading entry of the row above it (results in a STEP like
shape for leading entries)
All entries in a column below the leading entry are zero

37. Denition
Reduced Echelon form (Row Echelon form, REF): A rectangular
matrix is of Echelon form (Reduced Row Echelon Form or RREF) if
The leading entry in each nonzero row is 1
Each leading 1 is the only nonzero element in its column.

38. Examples of REF
 

 3 1 4 0 

 
 
0 −2 0 4
 
 
 
 
 

 0 0 0 0 

 
 
0 0 0 0
 
 
 

 0 3 1 4 0 4 6 1 

 
 
0 0 7 3 0 5 6 −4
 
 
 
 
 

 0 0 0 0 0 5 6 −4


 
 
0 0 0 0 0 0 3 0
 
 

39. Examples of RREF (Leading elements are ones)
 

 1 0 4 0 

 
 
0 1 0 4
 
 
 
 
 

 0 0 0 0 

 
 
0 0 0 0
 
 
 

 0 1 0 4 0 0 0 1 

 
 
0 0 1 3 0 0 0 −4
 
 
 
 
 

 0 0 0 0 0 1 0 −4


 
 
0 0 0 0 0 0 1 0
 
 