matrices and row operations

7/25/2019 Matrices and Row Operations

1/21

row

nmrows

=

mnmmm

n

n

n

aaaa

a

a

a

aaa

aaa

aaa

A

321

3

2

1

333231

232221

131211

A matrix is a rectangular array of

numbers. We subscript entries to telltheir location in the array

Matrices

are

identified

by theirsize.

Matrices and Row Operations

""

""

columnthjrowthiaij


2/21

14

51 [ ]20513

3

1

6

2

09749852

7531

4212

44


3/21

A matrix that has the same number of rows as

columns is called a square matrix.

maindiagonal

=

44434241

34333231

24232221

14131211

aaaa

aaaa

aaaa

aaaa

A


4/21

174242

3523

=+

=++

=+

zyx

zyx

zyx

=

741

412

523

ACoefficient matrix

If you have a system of

equations and just pick off

the coefficients and putthem in a matrix it is called a

coefficient matrix.


5/21

174242

3523

=+

=++

=+

zyx

zyx

zyx If you take the coefficientmatrix and then add a last

column ith the constants!it is called the augmented

matrix. "ften the constants

are separated ith a line.

=

1741

2412

3523

AAugmented matrix


6/21

We are going to work with our augmented matrix to get it in aform that will tell us the solutions to the system of equations.

The three things above are the only things we can do to the

matrix but we can do them together (i.e. we can multiply a

row by something and add it to another row).

perations that can be performed without

altering the solution set of a linear system

!. "nterchange any two rows

#. $ultiply every element in a row by a non%ero constant&. Add elements of one row to corresponding

elements of another

row


7/21

#100

##10

###1

After we get the matrix to look like our goal' we put

the variables back in and use back substitution to

get the solutions.

We use elementary row operations to make the matrix look

like the one below. The signs ust mean there can be anynumber here***we don+t care what.


8/21

#100

##10

###1

#se ro operations to obtain

echelon form$

1762

353

12

=++

=++

=++

zyx

zyx

zyx

1762

3153

1121

The augmented matrix

Work on this column first.

%et the & and then use itas a 'tool( to get zeros

belo it ith ro

operations.

We alreadyhave the &

here e

need it.

We)ll take ro & and multiply it by

* and add to ro + to get a ,.

-he notation for this step is

*r& r+e rite it by the ro e

replace in the matrix /see next

screen0.


9/21

1762

0210

1121

1762

3153

1121

1762

0210

1121

3r1+ r

2

3r1 3 6 3 3

+ r2 3 5 1 3

0 1 2 0

1o e)ll use + times ro & added to ro * to get a , there.

2r1+ r

3

1520

0210

1121

2r1 2 4 2 2

+ r3 2 6 7 1

0 2 5 1

1o our first column is

like our goal.


10/21

1520

0210

1121

r2

2r2 0 2 4 0

+ r3 0 2 5 1

0 0 1 1

2r2+ r3

1520

0210

1121

We)ll use ro + ith the &

as a tool to get a , belo it

by multiplying it by + andadding to ro *

#100

##10

###1

1o e)ll move to the second

column and do ro operations to

get it to look like our goal.

We need a & in the second

ro second column so e)ll

multiply ro + by &

1100

0210

1121

the second column is

like e need it no


11/21

#100

##10

###1

1o e)ll move to the third

column and e see for our goal

e just need a & in the third ro

of the third column. We have it soe)ve achieved the goal and it)s

time for back substitution. We

put the variables and 2 signs

back in.

1100

0210

11213ubstitute ,& in for zin

second equation to find y

1

02

12

=

=+

=++

z

zy

zyx

x

column

y

column

z

column

equal signs ( ) 012 =+y

2=y

3ubstitute & in for zand + foryin first equation to findx.

( ) ( ) 1122 =++x

2=x

Solution i! "2 # 2 # 1$


12/21

Solution i! "2 # 2 # 1$

1762

353

12

=++

=++

=++

zyx

zyx

zyx

-his is the only /x ! y ! z0 that

make A44 -5677 equationstrue. 4et)s check it.

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) 1172622

312523

11222

=++

=++

=++ -hese are all true.

%eometrically this meanse have three planes that

intersect at a point! a

unique solution.


13/21

#100

#010

#001

This method requires no back substitution.

When you put the variables back in' you have

the solutions.

-o obtain reduced ro echelon form! you continue to domore ro operations to obtain the goal belo.


14/21

4et)s try this method on the

problem e just did. We take the

matrix e ended up ith hen

doing ro echelon form$ 1762

353

12

=++

=++

=++

zyx

zyx

zyx

4et)s get the , e need in

the second column by

using the second ro as

a tool.

+r+r&

1o e)ll use ro * as a tool to

ork on the third column to get

zeros above the &.

#100

#010

#001

1100

0210

1121

1100

0210

1301

+r*r+

*r*r&

1100

0210

2001

1100

2010

2001

1otice hen e put the variables

and 2 signs back in e have the

solution

1#2#2 === zyx


15/21

The process of reducing the augmented matrix to echelon

form or reduced echelon form' and the process of

manipulating the equations to eliminate variables' is called-

%aussian 7limination


16/21

4et)s try another one$

1237

0432

6223

=+

=+

=+

zyx

zyx

zyx-he augmented matrix$

1237

0432

6223

If e subtract the second

ro from the first e)ll get

the & e need for the first

column.

r& r+

We)ll no use ro & as our

tool to get ,)s belo it.

2r&r+

1237

04326211

1237

12850

6211

7r&r*

4316100

12850

6211

We have the first column

like our goal. "n the

next screen e)ll ork

on the next column.

#100

##10

###1


17/21

1237

0432

6223

=+

=+

=+

zyx

zyx

zyx

If e multiply the second

ro by a &89 e)ll get the

one e need in the second

column.

We)ll no use ro + as our

tool to get ,)s belo it.

1/5r+

10r+r*

4316100

12850

6211

Wait: If you put

variables and 2 signs

back in the bottom

equation is , 2 &;

a false statement:

4316100

512

5810

6211

19000

5

12

5

810

6211

#100

##10

###1

I1


18/21

534

132

465

=

=+

=+

zyx

zyx

zyx

5134

1132

4165 "ne more$

#100

##10

###1

r& r*

2r&r+

4r&r*

5134

1132

1231

9990

3330

1231

1/3r+

9990

1110

1231

9r+r*

0000

1110

1231

"ops===last ro ended up all zeros. >ut variables and 2

signs back in and get , 2 , hich is true. -his is the

dependent case. We)ll figure out solutions on next slide.

5134

1132

1231


19/21

0000

1110

1231

x y z

zz

zy

zx

=

+=

+=

1

2

zz

zy

zx

=

=

=

1

2

0000

1110

2101

4et)s go one step further and get a ,

above the & in the second column

3r+r&

Infinitely many solutions here zis any real number

1o restriction on z

put variables

back in

solve forx? y


20/21

zz

zy

zx

=

+=

+=

1

2

Infinitely many solutions here zis any real number

534

132

465

=

=+

=+

zyx

zyx

zyx

What this means is that you can

choose any real number for zand

put it in to get thexand ythat goith it and these ill solve the

equation. @ou ill get as many

solutions as there are values of zto

put in /infinitely many0.

4et)s try z2 &. -hen y2 + andx2 *

( ) ( )

( ) ( )

( ) ( ) 512334

112332

412635

=

=+

=+orks in all *

4et)s try z2 ,. -hen y2 & andx2 +

( ) ( )

( ) ( )

( ) ( ) 501324

101322

401625

=

=+

=+

-he solution can be

ritten$ /z + ! z & !

z0


21/21

Acknoledgement

" wish to thank hawna /aider from alt 0ake 1ommunity 1ollege' 2tah2A for her hard work in creating this 3ower3oint.

www.slcc.edu

hawna has kindly given permission for this resource to be downloaded

from www.mathxtc.comand for it to be modified to suit the WesternAustralian $athematics 1urriculum.

tephen 1orcoran

/ead of $athematics

t tephen+s chool 4 1arramar

www.ststephens.wa.edu.au
http://www.slcc.edu/http://www.mathxtc.com/http://www.ststephens.wa.edu.au/http://www.ststephens.wa.edu.au/http://www.mathxtc.com/http://www.slcc.edu/

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