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Lecture notes Mathematics II (MATH2011A) - ALGEBRA_Chapter_3 R275,00
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Lecture notes Mathematics II (MATH2011A) - ALGEBRA_Chapter_3
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This document clearly describes, with detailed notes and examples, how to evaluate/solve the following: ~ Linear Spaces ~ Bases ~ Dimensions ~ Independence ~ Rank ~ Eigenvalues and Eigenvectors ~ Diagonalization ~ The Characteristic Polynomial as taught by the University of the Witwater...

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  • February 14, 2022
  • 35
  • 2021/2022
  • Class notes
  • Alexander davison
  • All classes

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CHAPTER 3 : LINEAR ALGEBRA

3. 1. Revision : MATRICES :


• GAUSS -
JORDAN ELIMINATION :


EXAMPLE :



SOLVE THE SYSTEM OF LINEAR EQUATIONS

I
x +
iy + 3iz + w =



ix ( Iti )z 2-i
-2g + + w =



Zix C-Zti)z Citi )w 2
3y + +
-
=




UNKNOWNS :




HERE
,
{ XIY ;ZjW} ARE COMPLEX NUMBERS AND COULD CALL

THEM { Zijzz ;Z3;Z4 }

I WRITE DOWN AN AUGMENTED MATRIX :

CAN SEE HAVE 3 EQNS AND 4 UNKNOWNS i. MORE UNKNOWNS THAN EQUATIONS
IN SOLUTION (POSSIBLY MORE ) :
'

- -
EXPECT 1 PARAMETER
,
MATRIX THAT


q
% "°WS LEAVE ROW 9- AS IS : SINCE 9- AS FIRST ENTRY : IDENTITY MATRIX !

3i I 1- 3i
~
9- i 1 i I 1


n.eu#0-14ti1-i22iRz-iR12i-3-2tilti
i -2 Iti I 2 -
i
2 0 -
I 4 +i 1- i 2- Zi Rs -
2iR1

-
IN GAUSS JORDAN ELIMINATION :


GOAL IS TO PERFORM ROW OPERATIONS
UNTIL LHS LOOKS AS CLOSE AS
POSSIBLE 70 IDENTITY MATRIX !
* CAN ADD SUBTRACT TWO ROWS
OR MULTIPLY A ROW BY CONSTANT /
COMPLEX CONSTANT



!

Ng R
'S AFTER 1
moa-n.us JORDAN WANT ZERO




}
~
I 0 7- i -
I Zti 3 1- Zi , + i. Rz
CANT MAKE IT
MORE
look ANY MORE CAN MAKE

0 I -4 j -
j -
f 2 I -2
-

R2
THAN IDENTITY
☐
g-
But
and
messes
o.si .


UP
"
MATRIX ACHIEVED
O
WHAT WAS

O O O O R2 -

R3 ALREADY !


µgIN GAUSS JORDAN ELIMINATION :


GOAL IS TO PERFORM ROW OPERATIONS
UNTIL LHS LOOKS AS CLOSE AS
POSSIBLE 70 IDENTITY MATRIX !



2 i. AUGMENTED MATRIX REPRESENTS 3 EQUATIONS :


ox
toy 1- Oz tow =o

USEFUL EQUATIONS :


K + C- 1t7i)z + (zti )w = 3t2i

y
t C- 4- i)z + ( i 1) w -
= Zi -2



° ⇐ AND y) o
-5
VARIABLES :(ELIMINATION
WAY PERFORM Gauss

} USE TWO EQUATIONS TO ELIMINATE ANY
-




FROM
EASY TO ELIMINATE
EQUATIONS !
X
Y

✗ = 3t2i -

C- lt7i)z (zti )w
y = Zi -2 -
C- 4- iz -

( i 1) w -

, 4 THE GENERAL SOLUTION :

4 UNKNOWNS :


→ VECTOR
K 3t2i + (I -
7-i)z 1-
f- 2- i)w
y -2+2 i + ( 4+i)z + ( I -
i )w
= * sum




{
Z £
DO NOT
HAVE ANY 2- sqYI.net
OTHER EQNS > NUMBER
W W
'

- - CANNOT ELIMINATE
Z AND W




SUM CONSTANTS

ors
3t2i 1- 7- i -
2- I
= '


-2+21 4ti 1- i
+ z + w
0 I 0

0 0 9-


-1--0
| IN SOLUTION :
Z AND W
p CAN PUT SEPARATELY
INTO SYSTEMS OF EQNS
/ COMBINED

FROM CALCULUS : BECOME ANYTHING
ARBITRARY CONSTANTS
-
EG . 2C = ( I 7- i)z + ( 2-I)w
- -



IS PARTICULAR -

PARAMETERS CORRECT RHS

(
NOT
1
SOLUTION OF ORIGINAL y Rµg = 0 LIKE PARTICULAR SOLNO
SYSTEM OF EQUATIONS
( No ARBITRARY constants :)
To SOLVE HOMOGENEOUS EQUATION !
IF SUBSTITUTE THESE ELEMENT OF NULL
CONSTANTS INTO EQNS SPACE CEE
= RHS

°
GOING FORWARD
,
COMMON FOR EQUATION TO BE HOMOGENEOUS ONLY !
( PARTICULAR SOLN = 0 )

•
TWO WAYS TO INVERT A MATRIX :( INVERSE OF MATRIX



EXAMPLE :

I Iti 9-
FIND THE INVERSE OF
i
Iti o

i i Iti


1 PERFORM GAUSS -
JORDAN ELIMINATION OR

2 WORK OUT DETERMINANT AND ADJOINT


I 1 WRITE AUGMENTED MATRIX :




I 9- + i 9- I o o ~ I Iti 1 1 00
Iti i 0 0 TO 0 -
i -
I -
i -
I -
i g- o R2 -
Citi)R1
i i Iti oE t R3
- ¥ÉeÉ
O I I -
I 0 I
-

IRI
MATRIX WANT INVERSE
MATRIX ON RHS
OF ON LHS !
GAUS / AN ELIMINATION
-
. . PERFORM
TO GET IDENTITY MATRIX ON
LHS AND INVERSE ON RHS !


I 9- ti I l 00 Iti 1
~ ~ I 1 00
o I 1- i 1- i i 0 Rzci) o I I -
i 1- i i o
O I I -
I 0 I 0 0 I -
I -
i I Rs -

Rz


~ I Iti I l 0 O ~ I 9-
' '
0 Ricki
Iti Iti 0
O I 1- i 1- iio o I 1- i 1- iio
O O I I -
I -
i R3( i ) -



O O I I -
I -
I Tako
not use
# FRACTIONS !

GO COLUMN BY COLUMN
WITH 1 'S FIRST THEN
ZERO 's !

, ' l

R1(¥+
-




Éii
'




i
→ it '
I 1
I
~ 9- 0 /+ i
-
0 R, -
Rz ~ i -
ai 0
iii. ,



0 I 1- i 1- ii o o I -
i 1- ii o

o o 1 i -
i -
i 0 0 1 i -
i -
i

? Iai ? %
-




~ I 0 0 i Ri -
R}
~
I 0 0 ' i
o I I -
i 1- i i o o I I 9- Ii 0 Rz(
0 0 1 i -
i -
i o o 1 i -
i -
i

i
i-zi-iz.li
-
3-
~
I 0 0
0 I 0 1- iii. i Rz -
Rs
o o 1 i -
i -
i
-
3- i i
1- Zi 1- zi I "
W"
i. INVERSE : !
1- i ¥ i approach
fractions
i -
i -
i no

L




z prEFERABLEMET

"Ñi^=detA%d¥a
:




¥É=
TO GET COFACTOR OF EACH ENTRY :



I +I
# BLOCK
9- 9- OUT ROW AND COLUMN OF POSITION IN
!
T
AND THEN WORK OUT DETERMINANT OF REMAINDER

'
1 A 2 iclti ) Ii)(Iti) E
i
adj A- ( Iti)
=
Iti 0 = -




i i Iti '
( Iti) -
i ( ai) i -
i -
( Hilli)
'
i-
-
( Iti) i -

( Iti )

+
I t
-
t -




t -
t

, iclti )
' T
2
adj A- = ( Iti) Ii)(lti) iz-



'
( Iti) -
i ( ai) i -
i -
( Hilli)
?
i -
-
( Iti) i -

( Iti )

ti -
l Zi + i T
adjA
-




-
i + I I -




+ -
i -
-
l -
it i -




:} ¥ÉE%umNs!
"
HAVE
STILL
T
→
TAKE TRANSPOSE yo

I -
I -
2J [ TO
of This
MATRIX
g,,
n.gg , ,n
,



adj A =
-
[ I -
I
co -




EACH
FACTOR
ENTRY
OF

!
-
i 9- + i -
i


i -
i -
i -
i
%
ddjA = -
Zi 1 9- ti

i-1•
METHODS
OUT
TO
DETERMINANT
WORK

9- + i I
i'
I • '

3 det A =
det i
g- + i 0

i i Iti det =
iclti) to + iciti ) -
iz - o -
( Iti )
's



i Zi Citi)
=
i -
i + -
i ti -




SINCE KNOW ADJOINT OF A 8 Zi l Zi 2i2
3 = -
-
-




= 9-
•

def A = 1st ENTRY IN A adj A 2 COULD EXPAND ON Row /column : THAT HAS ZERO IN IT !
A adjA
✗
:±①
+ -
+
=


I 9- + i 9- i -
i i -
i Iti i ti
? Yi
-

=
+1 det - Odet + 9-
+ idet 1 '


= Iti i 0 -
Zi 1 Iti
i i Hii

i i Iti i -
I -
i = i -1+1 -0 + ( Iti ) ( i - zi )
ziz )
'
=
i + (i ziti
3×3 3×3
- -




=
it i -

Zi -1+2
i. 3×3 = 9-

9-( i )
•


=
i -
i + ( Iti )( -
Zi) t • First
ENTRY !


-

= i -
i -
Zi -
Zi + i

= 1
:
4 A
-

= adjA
detA
i i i
÷
-
i -
-




=
-
zi I 9- ti
i -
I -
i

÷ A- = i -
I -
i -
i
-
Zi 1 Iti
i -
I -
i
0

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