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Section 1.2

Row Reductionand Echelon Forms

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Row Reduction

and Echelon Forms: e process o s mp y ng a ma r x us ng

row operations is called row reduction.

n s sec on, we re ne e me o o ec on . n o

a row reduction algorithm that will enable us to.

The algorithm we will discuss is known as Gaussian

.

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- -

a non-zero entry in that column. Interchange rows ifnecessary so that this entry is in the first row.

Step 2: Divide the top row by the non-zero entry from

Step 1, there should now be a 1 in the top left entry.This is called a leading one or a pivot.

Step 3: Eliminate all the entries beneath the pivot (in

the same column) by adding or subtracting multiples ofthe top row.

Step 4: Cover up the top row and repeat Steps 1through 3.

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se auss an e m na on o n a s mp er equ va en

system of linear equations. If possible, find the

.

=++ 92xxx

=+ 1342 321 xxx

= 321 xxx

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se auss an e m na on o n a s mp er equ va en

system of linear equations. If possible, find the

.

=

1 2 32 4 2 10 x x x

+ =

1 2 35 8 3 1 x x x =

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: ma r x s sa o e n row ec e onform if it satisfies the following three criteria:

. -leading one).

.bottom of the matrix.

in the row above it.

matrix, the end result will be in row echelon form.

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se auss an e m na on o row re uce e o ow ng

matrix. Check that resulting row equivalent matrix is in

.

0032

2134

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c o e o ow ng ma r ces are n row ec e on

form? If they are not, use Gaussian elimination to find

.

1111 000

2010

0010(A)

100

121(B)

2210 21

0100

40

(D)

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Uniqueness of

Row Echelon Form: s ng auss an e m na on o row re uce a

matrix does not always result in the same row echelon

1111

. ,

necessarily unique.

2010

0010e.g. The matrix can be row reduced to either

1111

1111

1000

1000

.

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-

n or er o row re uce a ma r x a s n row ec e on

form even further, we add the following additional

Step 5: Beginning with the right-most leading one, use-

terms above that leading one. Repeat for each leading

one, as you move from right to left across the matrix.

With this additional step, this row reduction algorithm

is known as Gauss-Jordan elimination.

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se auss- or an e m na on o ur er s mp y e

following matrix, which is already in row echelon form.

23100

02131

21000

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: ma r x s n re uce rowechelon form if it satisfies the following criteria:

. e ma r x s n row ec e on orm.

2. Each leading one is the only non-zero entry in itsco umn.

There are many different sequences of row operations

which can be applied to a matrix to find its RREF.However, using Gauss-Jordan Elimination always leads

to t e o t e matr x.

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e erm ne w e er or no eac o ese row ec e on

form matrices are in reduced row echelon form.

1 2 0 0 0 1 0 1 0 1 0 0 1 3 0

0 0 0 1 0

(A) (B) 0 1 5 0 2

0 0 0 1 3

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Uniqueness of the

Reduced Row Echelon Form,

row echelon form matrix which can be obtained from it.

That is, the reduced row echelon form of a matrix is

.

Thus, we refer to it as the reduced row echelon form

o t e matr x.

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se auss- or an e m na on o n e o e

following matrix.

410

121

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se auss- or an e m na on o n e so u on s o

the following system of equations.

1 35 1x x =

2 3

1 2 32 3 13 10 x x x

=

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Basic Variables

and Free Variables

Given a system of linear equations and its

,

column in the coefficient part of its RREF that has no

leadin one.

A free variable is a variable associated with a free-

Example: In Exercise 8, x1

and x2

are basic variables,

3.

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,

find the RREF of its augmented matrix. Determinewhich variables in the s stem are free and which are

basic.

= (B) =+ 62 321 xxx

=++ 42 321

321

xxx

= 12x = 13x

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Existence and Uniqueness

Theorem

the row echelon form or row reduced echelon form ofthe au mented matrix has no row of the form

.[ ]0 0 0 1

If a system of linear equations is consistent, then the

solution set contains either (i) a unique solution, when

ere are no ree var a es, or n n e y manysolutions, when there is at least one free variable.

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Using Row Reduction to

Solve a Linear System

Step 2: Use Gaussian elimination to obtain the row

Otherwise, go to Step 3.

Ste 3: Continue row reduction to obtain the rowreduced echelon form of the augmented matrix.

Ste 4: Write the s stem of e uations corres ondin

to the matrix obtained in Step 3.

Ste 5: Rewrite each non-zero e uation from Ste 4 sothat its one basic variable is expressed in terms of any

free variables appearing in the equation.

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using row reduction.

4321

= xxxx

=++ 22 321 xxx

(B)

=+

=+

0

62321

xx

xxx

(C) =+ 621 xx

=+ 522 21 xx