Gauss Jordan Method

(1)

(2)

Linear

Linear Sy

Sy

stems

y

Solve Ax=b, where A is an

nn

v

nn mm

atrix and

b is an

nn

v

₁₁

_colu

_m_m

_{n vector}

y

y CC

an also talk about non-

an also talk about non-square syste

square syste

mm

s where

A is

mm

v

nn

_{, b is}

mm

v

₁₁

_{, and x is}

nn

v

₁₁ y

y Overdetermined Overdetermined if if mm>>nn::



mmore equations than unknownsore equations than unknowns

y

y U U nderdetermined nderdetermined if if nn>>mm::



mmore unknowns than equationsore unknowns than equations

C

(3)

Singular Systems

y

A is singular if so

mm

e row is

linear co

mm

binatio

bination

n

of

of other rows

other rows

y

Singul

Singular

ar

syste

mm

s can be underdeter

mm

ined:

or inconsistent:

(4)

G

auss-Jordan Elimination

y

y FF

unda

mm

ental operations:

1.

1. R R eplace one equation with lineaeplace one equation with linear cr coommbinationbination

of other equations

2

2.. IInterchange two equationsnterchange two equations 3

3.. R R e-label two variablese-label two variables

y

y CC

o

mm

bine to reduce to trivial syste

mm y

y

Si

mm

plest variant only uses #

11

operations,

but get better

stabil

ity by adding

#2 (partial pivoting) or #2 and #3 (

(5)

G

auss-Jordan Elimination

y

Solve:

y

Only care about nu

mm

be

bers

rs



f

for

or

mm

tableau or

aug

(6)

G

auss-Jordan Elimination

y

y GG

iven:

y

y GG

oal: reduce this to trivial syste

mm

and read off a

(7)

G

auss-Jordan Elimination

y

y BB

asic operation

11

: replace any row by

linear co

mm

binatio

bination with any other r

n with any other row

ow

y

(8)

G

auss-Jordan Elimination

y

y R R

eplace

ro

row2 w

w2 w

ith r

ith row2 

ow2 

4 * row

11

y

(9)

G

auss-Jordan Elimination

y

y R R

epla

eplace ro

ce ro

w

11

with row

11



33//₂₂

* row2

y

(10)

G

auss-Jordan Elimination

y

y FF

or each row i:

y

y Multiply row i by Multiply row i by 1/1/aa_ii_ii y

y FFor each other row j:or each other row j:

y

y Add a Add a_ji_ji titimmes row i to row jes row i to row j

y

At the end, left part of

mm

atrix is

atrix is identity

identity

,

answer in right part

y

(11)

R

R ecall that we'd like to use row operations on an augecall that we'd like to use row operations on an augmmentedented

m

m_{atrix to get it into the following for}_{atrix to get it into the following for}mm_:_:

This is not always possibl

This is not always possible thoughe though.. The following areThe following are

m

matrices that cannot be put into this foratrices that cannot be put into this form.m.

1 1 2 2 3 3 1 1 1 1 00 00 00 00 0 0 11 00 00 00 0 0 00 11 00 00 0 0 00 00 11 00 0 0 00 00 00 11 n n n n b b b b b b b b b b   « « »» ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ - - ½½ L L M M O O MM L L L L 1 1 2 2 3 3 77 1 1 0 0 5 5 22 0 0 0 0 0 0 00 0 0 1 1 6 6 33 0 0 0 0 0 0 00 « « »» « « »» ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ - - ½½ ¬ ¬ ¼¼ - - ½½

(12)

R

R ecognize that if we cant get ourecognize that if we cant get our mmatrix to the desired foratrix to the desired formm,,

then it wont be as easy to see what the solution to the

syste

systemm _{of equations will be}_{of equations will be}..

F

For exaor exammple, thisple, this mmatrix has a solution that is easy atrix has a solution that is easy

to

to see, see, ((11, 3, 5), b, 3, 5), because tecause thehe mmatrix is in the finalatrix is in the final

for

formm that we wantthat we want..

1 1 0 0 0 0 11 0 0 1 1 0 0 33 0 0 0 0 1 1 55

«

»»

¬

¼¼

¬

¼¼

¬

¼¼

-

½½

(13)

1 1 2 2 3 3 77 0 0 0 0 0 0 00 0 0 0 0 0 0 00

« «

»»

¬ ¬

¼¼

¬ ¬

¼¼

¬ ¬

¼¼

- -

½½

1 1 0

0 5

5 2

2

0 0 1

1 6

6 3

3 « «

»»

¬ ¬

¼¼

- -

½½

This

This mmatrix (on the right) has a solution but isatrix (on the right) has a solution but is

not as clear what the solution is

not as clear what the solution is.. _{What we can}_{What we can}

conclude about the solution, (

conclude about the solution, ( x x,, y y,, zz), ), is is thatthat

the co

the commponentsponents_x_x,,_y_y, and, and _z_z mmust obey theust obey the

equation

equation x x + 2+ 2 y y + 3+ 3zz = 7= 7..

This

This mmatrix (on the right) has a solution, butatrix (on the right) has a solution, but

again it is not as clear what it is

again it is not as clear what it is.. What we canWhat we can

conclude about the solution, (

conclude about the solution, ( x x,, y y,, zz), ), is is thatthat

the co

the commponentsponents_x_x,,_y_y, and, and _z_z mmust obey the twoust obey the two

equations

(14)

These last two

These last two mmatrices represent systeatrices represent systemms that do nos that do not have at have a

unique solution

unique solution.. Whenever aWhenever a mmatrix does not have a uniqueatrix does not have a unique

solution (if it has inf

solution (if it has infinitelinitely y mm_{any solutions or no solution at all)}_{any solutions or no solution at all)}

we will not be able to get our aug

we will not be able to get our augmm_ented_ented mm_{atrix into the for}_{atrix into the for}mm

that we really want

that we really want.. When this happens, we want to at least getWhen this happens, we want to at least get

our

our mm_{atrix as close as possib}_{atrix as close as possib}_{le to}_{le to}_{this for}_{this for}mm _{that we would really}_{that we would really}

like it to be in

like it to be in.. _{When it}_{When it}_{is as close as it can pos}_{is as close as it can pos}_{sibly get, we}_{sibly get, we}_say_say

it is in reduced row echelon for

it is in reduced row echelon form.m.

1 1 2 2 3 3 1 1 1 1 00 00 00 00 0 0 11 00 00 00 0 0 00 11 00 00 0 0 00 00 11 00 0 0 00 00 00 11 n n n n b b b b b b b b b b   « « »» ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ ¬ ¬ ¼¼ - - ½½ L L M M O O MM L L L L