systems of linear equations

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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