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Announcements
Ï If anyone has not been able to access the class website pleaseemail me at [email protected]
Ï If anyone didn't get the email sent via the class listserv, pleasesee me.
Ï Homeworks are posted on the class website. I have made a fewchanges, so please make sure you have the updated list.
From Yesterday
Remember the 3 equations with 3 variables from yesterday?
x −2y +z = 0
3y −12z = 12
−8x +10y +18z =−18
Ï We want a solution (values of x , y and z) that will satisfy all 3equations.
Ï We will use matrix notation that we saw yesterday
Ï Start with the augmented matrix
Augmented matrix for the above system
1 −2 1 0
0 3 −12 12
−8 10 18 −18
Augmented column in circles
Then what?
Augmented matrix for the above system
1 −2 1 0
0 3 −12 12
−8 10 18 −18
Augmented column in circles
Then what?
Augmented matrix for the above system
1 −2 1 0
0 3 −12 12
−8 10 18 −18
We want the red numbers to be zero (as many as possible)
O� diagonal elements in the coe�cient matrix should become zero
Augmented matrix for the above system
1 −2 1 0
0 3 −12 12
−8 10 18 −18
We want the red numbers to be zero (as many as possible)
O� diagonal elements in the coe�cient matrix should become zero
To achieve this
Ï We do one or more of the 3 operations we discussed yesterday
Ï Add a row to a multiple of another row to replace that row(Choose the multiple so that it gives a zero somewhere)
Ï Multiply a particular row by a non-zero number
Ï Interchange any two rows.(Useful if the diagonal becomes zerosomehow)
Ï Copy the problem correctly. Do calculations carefully. Evensmall mistakes will lead you into wasting hours with nosolution in sight.
To achieve this
Ï We do one or more of the 3 operations we discussed yesterday
Ï Add a row to a multiple of another row to replace that row(Choose the multiple so that it gives a zero somewhere)
Ï Multiply a particular row by a non-zero number
Ï Interchange any two rows.(Useful if the diagonal becomes zerosomehow)
Ï Copy the problem correctly. Do calculations carefully. Evensmall mistakes will lead you into wasting hours with nosolution in sight.
To achieve this
Ï We do one or more of the 3 operations we discussed yesterday
Ï Add a row to a multiple of another row to replace that row(Choose the multiple so that it gives a zero somewhere)
Ï Multiply a particular row by a non-zero number
Ï Interchange any two rows.(Useful if the diagonal becomes zerosomehow)
Ï Copy the problem correctly. Do calculations carefully. Evensmall mistakes will lead you into wasting hours with nosolution in sight.
To achieve this
Ï We do one or more of the 3 operations we discussed yesterday
Ï Add a row to a multiple of another row to replace that row(Choose the multiple so that it gives a zero somewhere)
Ï Multiply a particular row by a non-zero number
Ï Interchange any two rows.(Useful if the diagonal becomes zerosomehow)
Ï Copy the problem correctly. Do calculations carefully. Evensmall mistakes will lead you into wasting hours with nosolution in sight.
To achieve this
Ï We do one or more of the 3 operations we discussed yesterday
Ï Add a row to a multiple of another row to replace that row(Choose the multiple so that it gives a zero somewhere)
Ï Multiply a particular row by a non-zero number
Ï Interchange any two rows.(Useful if the diagonal becomes zerosomehow)
Ï Copy the problem correctly. Do calculations carefully. Evensmall mistakes will lead you into wasting hours with nosolution in sight.
Ok. Let's do the row operations
1 −2 1 0
0 3 −12 12
−8 10 18 −18
8R1+
R3
Add 8 times �rst row to third row to get the "new" third row)
O� diagonal elements in the coe�cient matrix should become zero
Ok. Let's do the row operations
1 −2 1 0
0 3 −12 12
−8 10 18 −18
8R1+
R3
Add 8 times �rst row to third row to get the "new" third row)
O� diagonal elements in the coe�cient matrix should become zero
Result
1 −2 1 0
0 3 −12 12
0 −6 26 −18
Try not to lose existing zeros
Moving on
1 −2 1 0
0 3 −12 12
0 −6 26 −18
Divide second row by 3, makes life easier
This gives
1 −2 1 0
0 1 −4 4
0 −6 26 −18
Getting "ones" on the diagonal is good
Next
1 −2 1 0
0 1 −4 4
0 −6 26 −18
6R
2+R3
Add 6 times Row 2 to Row 3 to get "new" Row 3
Result
1 −2 1 0
0 1 −4 4
0 0 2 6
More zeros :-)
Next
1 −2 1 0
0 1 −4 4
0 0 1 3
Divide the last row by 2
Next
1 −2 1 0
0 1 0 16
0 0 1 3
R2+
4R3
Add Row 2 to 4 times Row 3 to get "new" Row 2
Next
1 −2 1 0
0 1 0 16
0 0 1 3
R1+
2R2
Add Row 1 to 2 times Row 2 to get "new" Row 1
Next
1 0 1 32
0 1 0 16
0 0 1 3
Almost there
Next
1 0 1 32
0 1 0 16
0 0 1 3
R1-R
3
Do Row 1 - Row 3 to get "new" Row 1
Next
1 0 0 29
0 1 0 16
0 0 1 3
Bingo!!
The solution is staring right at us
The coe�cient matrix is now a TRIANGULAR matrix
Any preferences? Order?
Ï None! You could do any of these operations in any order youwant
Ï You could combine two or more operations in one step as youpractice more problems.
Ï If your matrix is getting worse with each step, make sure youcopied the right problem and check your calculations.
Another example
x +y +z = 3
2x −y −z = 5
2x +2y +2z = 7
1 1 1 3;
2 −1 −1 5;
2 2 2 7;
Vertical line separates augmented column
Row Operation
1 1 1 3
2 −1 −1 5
0 0 0 1
R1-R
3
R3 - 2R1 to give "new" Row 3
We have a problem, 0=1!!!!!
Row Operation
1 1 1 3
2 −1 −1 5
0 0 0 1
R1-R
3
R3 - 2R1 to give "new" Row 3
We have a problem, 0=1!!!!!
Inconsistent System
The above example is an inconsistent system. In other wordswhenever your row reduced matrix looks like (could happen in anyrow)
a b c d
0 f g h
0 0 0 ∗
Here "*" is a non-zero number
Inconsistent System
Ï If all elements in a row left to the augmented column are zerowith a non-zero element in the augmented column, the systemis inconsistent (no solutions, parallel planes)
Ï Usually happens when an equation is multiplied by a certainnumber 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
Problem 12 sec 1.1
x −3y +4z =−43x −7y +7z =−8−4x +6y −z = 7
1 −3 4 −4
3 −7 7 −8
−4 6 −1 7
Vertical line separates augmented column
Problem 12 sec 1.1
1 −3 4 −4
0 2 −5 4
0 −6 15 −9
R2-3R1 for "new" R2 and R3+4R1 for "new" R3
Problem 12 sec 1.1
1 −3 4 −4
0 2 −5 4
0 0 0 1
Do 1
3R3 and add to R2 to get new R3
Inconsistent
Problem 20 sec 1.1
Determine the value of h so that the following is the augmentedmatrix of a consistent linear system.
2 −3 h
−6 9 5
Solution: Add 3R1 to R2 to get new R2 (Don't forget that theaugmented 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.
Sec 1.2, Row reduction, Echelon forms
To develop an e�cient algorithm for any matrix irrespective ofwhether it represents a linear system.
Ï Nonzero row/column means a certain row/column has atleastone nonzero entry
Ï Leading entry of a row means the �rst nonzero entry in a row(left most)
De�nition
Echelon form (Row Echelon form, REF): A rectangular matrix is ofEchelon form (Row Echelon Form or REF) if
Ï All nonzero rows are ABOVE any rows with all zeros
Ï Each leading entry of a row in a column is to the RIGHT tothe leading entry of the row above it (results in a STEP likeshape for leading entries)
Ï All entries in a column below the leading entry are zero
De�nition
Reduced Echelon form (Row Echelon form, REF): A rectangularmatrix is of Echelon form (Reduced Row Echelon Form or RREF) if
Ï The leading entry in each nonzero row is 1
Ï Each leading 1 is the only nonzero element in its column.
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
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