Row Reducing Matrices A Presentation on Procedures Used to Work with Matrices East Los Angeles...

Row Reducing Matrices

A Presentation on

Procedures Used to

Work with Matrices

East Los Angeles CollegeMEnTe Program

David Morín

Prepared by

EXITTOPICS BACK NEXT


IntroductionRow OperationsRow Echelon FormReduced Row Echelon FormCheck ColumnCross Product MethodFinding the Inverse of a Matrix

Topics

Click on a link to go to that slide or press the left mouse button to go the next slide.



The process of row reducing a matrix is used for working with an augmented matrix. The augmented matrix may represent a system of equations to be solved, or the augmented matrix may be a matrix combined with the identity matrix. We will use the latter case when we work on finding the inverse of a matrix.



Row Operations

The process of row reducing matrices involves three basic row operations.

1. Interchange any two rows

2. Replace a row by a non-zero multiple of that row

3. Replace a row by the sum of that row and a constant non-zero multiple of some other row

We will next illustrate each of these operations.EXITTOPICS BACK NEXT


We can interchange any two rows. For Example,

2 1 3

3 1 6

4 2 5

2 1 3

4 2 5

3 1 6

becomes

if we interchange rows two and three.



We can replace any row by a non-zero multiple of that row. For Example,

4 2 6

4 2 5

9 3 18

2 1 3

4 2 5

3 1 6

becomes

if we multiply the top row by 2 and the bottom row by –3.



We can replace a row by the sum of that row and a constant non-zero multiple of some other row. For example,

2 1 3

10 0 17

3 1 6

2 1 3

4 2 5

3 1 6

becomes

by replacing row two with the sum of row two plus 2 times row three.



Elementary row operations on a matrix are usually performed with the intention of simplifying the matrix. A matrix is simplified by reducing as many elements as possible to zeros.

There are two basic simplifications of a matrix.

The Purpose of Row Operations



If a row does not consist entirely of zeros, then the leading entry is a 1.Any rows that consist entirely of zeros are grouped together at the bottom.In rows that do not consist entirely of zeros, the leading 1 in lower rows occur farther to the right than the leading 1 in higher rows.

The first simplification of a matrix is to put the matrix in row-echelon, or triangularized form. In this form the following properties must be true.

Row-Echelon Form

•

•

•



A Matrix Reduced to Row-Echelon Form

2 1 3 10

1 2 1 3

2 1 2 3

can be reduced to

1 2 1 3

0 1 0 3

0 0 1 1

The resulting matrix is in row-echelon form. The next few slides explain how to achieve this result.



2 1 3 10

1 2 1 3

2 1 2 3

1 2 1 3

2 1 3 10

2 1 2 3

becomes

This is done by interchanging rows one and two (whenever possible, it is preferable to have a 1 as the first element in the first row).



1 2 1 3

2 1 3 10

2 1 2 3

1 2 1 3

0 5 1 16

0 3 0 9

becomes

This is the result of replacing row two with row two plus -2 times row one and also replacing row three with row three plus -2 times row one. This reduces the first element of rows two and three to zero and results in having the leading element in these rows to the right of the leading element in row one.



1 2 1 3

0 5 1 16

0 3 0 9

1 2 1 3

0 1 0 3

0 5 1 16

This is the result of interchanging rows two and three and then multiplying the new row two by (or dividing by 3). Again, we prefer to have a 1 in the leading element.

13

becomes



1 2 1 3

0 1 0 3

0 5 1 16

1 2 1 3

0 1 0 3

0 0 1 1

becomes

And finally, this is the result of replacing row three with row three plus –5 times row two.

Notice that the bottom left corner of the matrix forms a triangle of zeros, hence, it is triangularized, or in row-echelon form.



We can also combine the second and third row operations to replace a row with the sum of a multiple of that row plus a multiple of another row.

If we wish, rather than taking the sum of a multiple of a row plus a negative multiple of another row, we can take the difference of a multiple of a row minus a multiple of another row.

Expanding the Row Operations



A second way to simplify is to put the matrix in reduced row-echelon form. Simplifying this way is also know as using the Gauss-Jordan method.

The reduced row-echelon form is an extension of the row-echelon form. It involves using the leading element in each row of the matrix to reduce all the other elements in that column to zeros, not just the elements below.

Reduced Row-Echelon Form



If we take the matrix that we had changed to row echelon form in the previous work, we can further reduce it by changing the remaining elements (except the last column) to zero as in the following example.

1 2 1 3

0 1 0 3

0 0 1 1

1 0 0 2

0 1 0 3

0 0 1 1

becomes

This is accomplished by adding 2 times row two to row one and subtracting 1 times row three from row one.



Helpful Hints in Row Reducing Matrices

• As pointed out before, if it is possible to make the leading element 1, do so by interchanging rows where necessary.

• Sometimes it is possible to divide a row by a common factor to make the leading element 1.

• If the row does not have a common factor, you can avoid dividing if you reduce a row by replacing it with a multiple of the row with the leading element plus a multiple of the row you are reducing. See the next slide for an example.



2 1 3 10

5 2 1 3

3 1 2 3

In the matrix below, none of the elements in the first column is 1, but we can use the first row to reduce row two by replacing row two with 2 times row two minus 5 times row one. In a similar manner, we can reduce row three by replacing it with 2 times row three minus 3 times row one.

2 1 3 10

0 9 13 56

0 5 5 24

becomes



The results we got in the last matrix, involves large numbers, but the only other alternative would have been to divide row one by 2, to make the leading element 1,but this wouldhave involved working with fractions, and the resulting matrix could have ended up more complicated than merely using large numbers.

Note that the matrix above can be somewhat simplified by multiplying rows two and three by –1 to eliminate the negative numbers.

2 1 3 10

0 9 13 56

0 5 5 24



Add another column to the right of the matrix. This will be the check column. The elements in the check column consist of the sum of the elements in each row of the matrix.

Using a Check Column



2 3 6 1

1 4 2 3

3 5 4 4

2 3 6 1 0

1 4 2 3 6

3 5 4 4 8

gets expanded

to

Adding a Check Column: An Example

Notice 2 3 ( 6) 1 0

1 4 ( 2) 3 6

3 5 ( 4) 4 8



Why a check column?

In row reducing matrices it is extremely easy to make mistakes. If the check column is set up as the sum of the elements in each row of the matrix, then as we use row operations the check column will continue to be the sum of the elements for each row.

When you are row reducing matrices, if the element in the check column ever turns out to be different from the sum of the elements in that row of the matrix, then there must be a mistake in the calculations for that row (the mistake may be in the check column, so check all elements of the row).



The Check Column

Here we will use an arrow to indicate the change from one matrix to its reduced form, and we add the check column to the right of each matrix.

2 1 3 10 6

1 2 1 3 3

2 1 2 3 6

1 2 1 3

2 1 3 10

2 1 2 3

3

6

6

If we interchange rows we also interchange the elements in the check column, so the correct sum still corresponds to each row.



1 2 1 3

2 1 3 10

2 1 2 3

3

16

6

3

22

12

1 2 1 3

0 5 1 16

0 3 0 9

3

12

22

1 2 1 3

0 3 0 9

0 5 1 16

1 2 1 3

0 1 0 3

0 5 1 16

3

4

22

Notice that as we reduce each row including the check column, the check column still gives the sum for each row.



1 2 1 3

0 1 0 3

0 5 1 16

3

4

22

1 2 1 3

0 1 0 3

0 0 1 1

3

4

2

Here we have row-echelon form.

1 0 0 2 3

0 1 0 3 4

0 0 1 1 2

Here we have reduced row-echelon form.

1 0 1 3 5

0 1 0 3 4

0 0 1 1 2




A technique that can be used to row reduce matrices is the technique of using cross products.

In Linear Programming, the leading element that is used to eliminate the remaining elements in a column is referred to as the pivot. We will use that term in this work.

We will use the pivot as one corner of a rectangle. The two pairs of opposite corners of this rectangle will be multiplied to find two products. We will then subtract these two products in much the same way we subtract the cross products in evaluating a 2 2 determinant.


Using Cross Products to Row Reduce Matrices



Look at the matrix below. If we use the first-row, first-column element as the pivot (denoted by the ), we can change any element in any other row and column by forming a rectangle whose corners include the pivot and the element we wish to change. If we want to change the –3 in the second-row, third-column, we get the rectangle shown on the right.

2 1 3 10

1 2 1 3

2 1 2 3

2 1 3 10

1 2 1 3

2 1 2 3



In using the cross product method, we take the product of the corners with the pivot minus the product of the remaining two corners,

(2)(-3) – (1)(10) = -16

We can use this method to change all the elements in the second row. Thus, we can change the –2 to (2)(-2) (1)(1) = 5, and the 1 in the third column changes to (2)(1) (1)(3) = 1.The 1 in the first column will change to a zero.

2 1 3 10

1 2 1 3

2 1 2 3



To understand what we are doing when we use the cross product method, you will need to realize that in this method, all the elements in row two are being multiplied by the pivot, the elements of row one are being multiplied by the element in the pivot column in row two and we are subtracting the

two rows. This is the same as replacing a multiple of row two with a multiple of row two minus a multiple of row one. It is the same as one of the row operations we would normally use to row reduce matrices. Take note that the element below the pivot will become a zero.

2 1 3 10

1 2 1 3

2 1 2 3


(1)(1) (2)( 2) 5

(1)(3) (2)(1) 1

(1)(10) (2)( 3) 16

(1)(16) (2)( 3) 22

2 1 3 10 16

1 2 1 3 3

2 1 2 3 6


We can now illustrate the cross-product method to completely row reduce a matrix. We will add a check column to verify our work. Reducing the second row, we get the following.

1 2 1 3 3

0 5 1 16 22

2 1 2 3 6

1 2 1 3 3

2 1 3 10 16

2 1 2 3 6



Now, using the same process, we can reduce the bottom row.

Using row operations, we can divide row three by 3 and interchange it with row two. This gives the following.

1 2 1 3 3

0 1 0 3 4

0 5 1 16 22

1 2 1 3 3

0 5 1 16 22

0 3 0 9 12

Now we are ready to work with a new pivot. The leading element in the second row is 1. This will be our new pivot.

1 2 1 3 3

0 5 1 16 22

2 1 2 3 6



We can use the new pivot to reduce the last row. This will give us a matrix in row-echelon form.

If we wish to change the matrix to reduced row-echelon form, we must also use this pivot to reduce the elements in the top row.

1 2 1 3 3

0 1 0 3 4

0 0 1 1 2

1 2 1 3 3

0 1 0 3 4

0 5 1 16 22


1 2 1 3 3

0 1 0 3 4

0 0 1 1 2

1 0 0 2 3

0 1 0 3 4

0 0 1 1 2

1 0 1 3 5

0 1 0 3 4

0 0 1 1 2


And finally, we use the leading element in the last row as the pivot and reduce the elements in rows one and two.

1 0 1 3 5

0 1 0 3 4

0 0 1 1 2



Notes on Using the Cross-product Method

1. As you use the pivot to reduce the other rows, all the other elements in the pivot column will be changed to zero.

2. Remember that the new values in the reduced row come from the cross product that includes the pivot minus the other cross product. This means you need to be careful with which cross product is subtracted from the other.





The Inverse of a Matrix

The inverse of an n n matrix, A, is the n n matrix, A-1, such that the matrix product, A·A-1 = A-1·A= I,where I is the n n identity matrix.

A non-square matrix cannot have an inverse, but it is also possible that a square matrix will not have an inverse.

If a square matrix has an inverse, one way to find the inverse is to use an augmented matrix composed of two halves. The left half will be the matrix whose inverse we are seeking and the right half will be the identity matrix.



The Inverse of a Matrix

To find the inverse matrix of the matrix below on the left we use the augmented matrix on the right.

2 1 3

1 2 1

2 1 2

2 1 3 1 0 0

1 2 1 0 1 0

2 1 2 0 0 1

To find the inverse, we will change the left side to the reduced row-echelon form by using row operations on the entire augmented matrix.


1 2 1 0 1 0 1

2 1 3 1 0 0 7

2 1 2 0 0 1 4


2 1 3 1 0 0 7

1 2 1 0 1 0 1

2 1 2 0 0 1 4

We will use row operations, along with the check column to row reduce the left side to the identity matrix.

1 2 1 0 1 0 1

0 5 1 1 2 0 5

0 3 0 0 2 1 2

1 2 1 0 1 0 1

0 3 0 0 2 1 2

0 5 1 1 2 0 5


3 0 3 0 1 2 7

0 3 0 0 2 1 2

0 0 3 3 4 5 5


9 0 0 9 15 21 6

0 9 0 0 6 3 6

0 0 3 3 4 5 5

1 2 1 0 1 0 1

0 3 0 0 2 1 2

0 5 1 1 2 0 5

Notice that because there was already a 0 in row two in the pivot column, we did not have to reduce it. The only change we got is that row two got multiplied by 3.



9 0 0 9 15 21 6

0 9 0 0 6 3 6

0 0 3 3 4 5 5

Finally, we can get the identity on the left by dividing each row by the leading element.

5 7 2

3 3 3

2 1 2

3 3 3

4 5 5

3 3 3

1 0 0 1

0 1 0 0

0 0 1 1



5 71

3 32 1

03 3

4 51

3 3

So the inverse of

2 1 3

1 2 1

2 1 2

is

3 5 71

0 2 13

3 4 5



2 1 3

1 2 1

2 1 2

To verify that we have the correct inverse matrix, we multiply the two to get the identity.

1 0 0

0 1 0

0 0 1

3 5 71

0 2 13

3 4 5

3 0 0

10 3 0

30 0 3










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