Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
Check:
Which a g r e e s w i t h ( 2 )
.
Note #1
Given s p e c i f i c numbers, a s i n t h i s e x e r c i s e , we would most
l i k e l y u s e ( 3 ) r a t h e r t h a n (1). The t e c h n i q u e used i n (1) i s
most u s e f u l when w e a r e working w i t h " l i t e r a l " c o n s t a n t s .
T h a t i s , s i n c e we cannot s i m p l i f y a 11 + bll, we would u s e (1)
t o obtain
[This i s analogous t o t h e s i t u a t i o n i n a l g e b r a where one l e a r n s
t h a t ( a + b ) 2 = a 2 + 2ab + b 2 , y e t computes ( 3 + 4 1 2 a s 72 = 49,
r a t h e r t h a n a s 32
+
2(3) ( 4 )
+
4 2 . ~
Note #2
The v a l i d i t y o f ( 4 ) s t e m s from t h e f a c t t h a t
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.1 (L) c o n t i n u e d
denotes v
D(v + v l t , v2) where
1
+
= allul
1 a12u2
v ' = b u + b12u2 1
11 1
V2
-
- a21U1
+
a22U2
and one o f o u r axioms f o r D i s t h a t
Note # 3
While o u r d e m o n s t r a t i o n was f o r n = 2, t h e r e s u l t i s v a l i d f o r
a l l n.
T h i s i s due t o t h e f a c t t h a t D i s d e f i n e d t o be l i n e a r .
That i s
f o r any s p a c e V = [ul,
...,un1
•
Note # 4
While w e s h a l l n o t b o t h e r t o prove it h e r e , t h e r e i s an i n t e r e s t i n g
r e s u l t which r e l a t e s t h e d e t e r m i n a n t of a m a t r i x t o t h e d e t e r minant o f t h e t r a n s p o s e o f t h e m a t r i x .
E s s e n t i a l l y , any
theorems r e f e r r i n g t o t h e rows o f a m a t r i x remains v a l i d f o r
the columns.
I n p a r t i c u l a r , t h e determinant of a matrix equals
t h e d e t e r m i n a n t of i t s t r a n s p o s e .
i d e a s i n p a r t (b)
.
W e s h a l l i l l u s t r a t e these
W e may u s e t h e t e c h n i q u e o f p a r t ( a ) s u c c e s s i v e l y t o o b t a i n
But
and
Solutions
Block 3: S e l e c t e d T o p i c s i n L i n e a r A l g e b r a
U n i t 5: D e t e r m i n a n t s
3.5.1 (L) c o n t i n u e d
Using ( 6 ) and ( 7 ) i n ( 5 ) w e o b t a i n
A s a check,
which a g r e e s w i t h ( 8 ) .
Note #5
W e c o u l d h a v e u s e d columns r a t h e r t h a n rows t o s o l v e t h i s problem
as f o l l o w s :
More l i t e r a l l y
Solutions
Block 3: S e l e c t e d T o p i c s i n L i n e a r Algebra
Unit 5: Determinants
3.5.1 (L) c o n t i n u e d
A s a check o f t h i s l a s t r e s u l t we may o b t a i n by d i r e c t compu-
tation:
On t h e o t h e r hand, u s i n g columns r a t h e r t h a n rows, we would
obtain
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.1 (L) c o n t i n u e d
The r i g h t s i d e s o f ( 9 ) and (10) a g r e e s i n c e t h e r i g h t s i d e of
Note #6
Equation ( 9 ) r e v e a l s a r e s u l t t h a t may s e e m a b i t " u n p l e a s a n t " .
Namely, it might seem n i c e t o b e l i e v e t h a t " t h e d e t e r m i n a n t of
a sum
e q u a l s t h e sum o f t h e d e t e r m i n a n t s " .
( 9 ) shows t h a t t h i s need n o t b e t r u e .
But, e q u a t i o n
I n more d e t a i l , suppose
we let
Then
Now, l e t t i n g [ A [ d e n o t e t h e d e t e r m i n a n t of A , e t c . , w e have
from ( 8 ) t h a t
Hence,
W e i l l u s t r a t e t h i s idea i n p a r t (c).
c.
W e have
and B =
Therefore,
r
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.1 (L) c o n t i n u e d
Hence,
while
IA + BI
T h i s i s s u f f i c i e n t t o prove t h a t
IAI +
need n o t e q u a l
14.
Note 97
I n t h i s e x e r c i s e w e saw t h a t [ A
+ BI - IAI
- IBI
= 15.
Accord-
i n g t o (11), t h e n
s h o u l d e q u a l 15.
A s a check,
Note # 8 Do n o t c o n f u s e p a r t s ( b ) and (c)
could n o t r e p l a c e
IA + BI
by
.
What ( c ) showed was t h a t w e I A ~ + IBI .
What (b) showed was I n o t h e r words, ( b ) t h a t w e could o b t a i n a " c o r r e c t i o n f a c t o r " .
showed t h a t w e c o u l d expand
+
by t a k i n g one row (column) IA
BI
a t a time. I n t h e p r e v i o u s e x e r c i s e we may have been a b i t crushed t o d i s c o v e r t h a t t h e d e t e r m i n a n t of a sum need n o t b e t h e sum o f t h e
determinants.
I f n o t h i n g else, such a r e s u l t s h o u l d a t l e a s t
make us a b i t wary of t r u s t i n g o u r i n t u i t i o n , and a c c o r d i n g l y ,
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.2 (L) c o n t i n u e d
it s h o u l d make us s e e m a b i t more t o l e r a n t when it comes t o
giving proofs
.
Our p o i n t i s t h a t one v e r y i m p o r t a n t p r o p e r t y o f d e t e r m i n a n t s i s
t h a t t h e d e t e r m i n a n t o f a p r o d u c t i s t h e p r o d u c t of t h e d e t e r minants; b u t w e cannot p a s s t h i s o f f a s b e i n g " s e l f - e v i d e n t " i f
o n l y because t h e c o r r e s p o n d i n g s t a t e m e n t about sums seems j u s t
as self-evident.
What w e s h a l l do i n t h i s e x e r c i s e i s prove t h e r e s u l t f o r t h e
c a s e n = 2. Our t e c h n i q u e s h a l l b e t h e one which g e n e r a l i z e s
t o a l l dimensions, b u t by r e s t r i c t i n g o u r a t t e n t i o n t o n = 2
we may a v o i d p a r t o f t h e mass of computational d e t a i l and
n o t a t i o n t h a t o f t e n o b s c u r e s t h e s t r u c t u r e of t h e p r o o f .
As
w e s h a l l see, t h e proof r e l i e s r a t h e r s t r o n g l y on t h e r e s u l t s
o f E x e r c i s e 3.5.1.
a.
Since
w e can b e c e r t a i n t h a t
By t h e l i n e a r p r o p e r t i e s of t h e d e t e r m i n a n t , a s d i s c u s s e d i n
E x e r c i s e 3.5.1, we see from (1) t h a t
may b e w r i t t e n a s
IAB~
~ a c t o r i n go u t t h e common f a c t o r s i n t h e rows o r columns o f t h e
t e r m s i n ( 2 ) , we s e e t h a t ( 2 ) may be r e w r i t t e n i n t h e form
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.2 ( L ) continued
But both
and
lb21
a r e 0 s i n c e each of t h e s e determinants has two e q u a l rows
[i.e.,
D(v v ) = 0 , e t c . ] .
1' 1
Moreover,
s i n c e one determinant i s o b t a i n e d from t h e o t h e r merely by i n t e r changing two columns [ i . e . ,
With t h e s e o b s e r v a t i o n s ,
i s a number n o t a m a t r i x ,
expression.
S.3.5.8
D(vlrv2) = -D(v2,vl) , e t c . l
(3) becomes
and can b e f a c t o r e d from t h e g i v e n
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.2 (L) c o n t i n u e d
Finally, since
w e see from (1) and ( 4 ) t h a t
Note #1
I n t h e r e l a t i v e l y s i m p l e c a s e n = 2 , w e c o u l d v e r i f y by d i r e c t
computation t h a t lABl = / A / l B l b u t n o t i c e how cumbersome t h e
d i r e c t computation becomes f o r l a r g e v a l u e s o f n.
Our t e c h n i q u e
n e v e r i n v o l v e s o u r h a v i n g t o expand by c o f a c t o r s , e t c . ,
but
r a t h e r h a s u s r e p e a t e d l y w r i t e a d e t e r m i n a n t a s a s i m p l e r sum
of d e t e r m i n a n t s .
Moreover, o u r t e c h n i q u e does n o t r e q u i r e
t h a t we c o n j e c t u r e t h e r e s u l t i n advance.
with
b.
1-1
and showed t h a t
S i n c e D(ul,
...,un)
IAB~
=
That i s , we began
IAIIBI.
= 1, o u r coding system s a y s t h a t
I n o t h e r words, t h e d e t e r m i n a n t of I i s 1.
NOW suppose A i s i n v e r t i b l e , t h a n A-le x i s t s such t h a t
Accordingly,
IAA-'~ =
111
.
(6)
S.3.5.9
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.2(L)
continued
By p a r t ( a ) w e know t h a t I Z U ~ ' ~ I=
called that
1II
= 1.
IAI
J A - ~ I and w e have j u s t r e -
Hence ( 6 ) becomes
From (7) w e o b t a i n
That i s
= 111-l.
Note #2
The b e a u t y of ( 8 ) i s t h a t it i d e n t i f i e s t h e i n v e r s e o f a
m a t r i x w i t h t h e i n v e r s e o f a number.
T h a t i s , IA-'I
refers t o
f i n d i n g t h e d e t e r m i n a n t o f t h e m a t r i x which i s t h e i n v e r s e o f A ,
w h i l e ] A 1-I r e f e r s t o f i n d i n g t h e d e t e r m i n a n t o f A (which i s a
number) and t h e n t a k i n g i t s r e c i p r o c a l .
p a r t (c)
c.
.
With
w e see a t once t h a t
W e may now f i n d A-'
, namely:
We illustrate this in
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.2 (L) c o n t i n u e d
hence,
Accordingly,
H e r e a r e a few p o i n t s which o c c u r j u s t w i t h i n t h e c o n f i n e s o f
the present exercise.
1.
The m a t r i x A-'
i n (10) can b e w r i t t e n a s
T h a t i s , when w e m u l t i p l y a m a t r i x by a number, each e n t r y o f
t h e m a t r i x i s m u l t i p l i e d by t h a t number.
On t h e o t h e r hand, when w e m u l t i p l y t h e d e t e r m i n a n t o f A-l,i. e . ,
~ A - ~, I by a number w e m u l t i p l y any one row ( o r column, b u t n o t
-1
For example, i n computing [ A 1
b o t h ) o f A-' by t h a t number.
from
(lo),
we find that
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
-
3.5.2 (L) c o n t i n u e d
- l
h
9
T
h
a
t i s , w e " f a c t o r e d o u t " 113 t w i c e , once f o r
each row ( o r column) i n which it was a common
factor.
Had w e simply w r i t t e n t h a t
we would have o b t a i n e d t h e i n c o r r e c t answer, 1, a s t h e v a l u e
of t h e determinant.
2. I f AB = 0 (where now 0 d e n o t e s t h e z e r o matrix), t h e n it
i s n o t t r u e t h a t e i t h e r A o r B must b e t h e z e r o matrix.
For
example,
but neither
nor
is t h e zero matrix.
T h i s s h o u l d n o t b e confused w i t h t h e numerical r e s u l t t h a t i f
IAB~
= 0 (where now 0 is t h e number 0) t h e n
IA I=
0 or
IBI
I n summary, t h e p r o d u c t of two m a t r i c e s can b e t h e z e r o .
= 0.
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.2 (L) c o n t i n u e d
Comparing (9) and (11) w e conclude t h a t
Note # 3
T h i s e x e r c i s e proves t h a t i f a m a t r i x i s i n v e r t i b l e , i t s d e t e r minant cannot e q u a l z e r o .
I A ~ nor 1
neither
More g e n e r a l l y , s i n c e
JBI
Namely, s i n c e
can b e z e r o .
IABI=
IAl
IBI , we
see that
# 0 - + + l ~ . I 3#l 0; o r from a d i f f e r e n t emphasis,
= 0 or
IBI
IA~# 0
i f IABI=
and
o++~A/
= 0.
Note # 4
Because of a l l our p r e v i o u s work w i t h m a t r i c e s and t h e r e l a t i v e l y
l i t t l e work we've done w i t h d e t e r m i n a n t s , t h e r e i s a danger t h a t
we may confuse m a t r i x p r o p e r t i e s w i t h d e t e r m i n a n t p r o p e r t i e s .
Many non-zero m a t r i c e s ( r e c a l l t h a t t h e zero-matrix i s t h e m a t r i x ,
a l l of whose e n t r i e s a r e z e r o ) have a z e r o d e t e r m i n a n t . I n f a c t ,
t h i s i s p r e c i s e l y t h e d e f i n i t i o n o f a s i n g u l a r m a t r i x . A t any
r a t e , n o t i c e t h a t t h e p r o d u c t o f two m a t r i c e s can be t h e zerom a t r i x even though n e i t h e r m a t r i x i s t h e zero-matrix.
For
example,
The two m a t r i c e s on t h e l e f t s i d e o f t h e e q u a l i t y a r e s i n g u l a r
b u t non-zero.
However, t h e p r o d u c t o f two m a t r i c e s cannot b e
s i n g u l a r u n l e s s a t l e a s t one o f t h e two m a t r i c e s i s s i n g u l a r .
I t i s t h i s f a c t t h a t i s s t a t e d i n terms of d e t e r m i n a n t s by
IAB~
= 0 i f and o n l y i f
IAJ
= 0 or
IBI
= 0.
S t a t e d w i t h o u t t h e language of d e t e r m i n a n t s , a l l we a r e s a y i n g
i s t h a t i f b o t h A-land B-le x i s t then ( A B ) - l a l s o e x i s t s , and
-1 -1
-lA-li n p a r t i c u l a r , it i s g i v e n by B A
[ s i n c e (AB)B
- A(BB-~)A-'
= *-I
= I].
S.3.5.13
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
-
--
3.5.2 (L) continued
I n summary, we must n o t confuse
the z e r o
m a t r i x with a m a t r i x
whose determinant i s z e r o ( a s examples of which a r e a l l s i n g u l a r
matrices)
.
3.5.3(L)
I f P i s an n by n i n v e r t i b l e m a t r i x and A i s any n by n matrix,
then we have:
[PI
The f a c t t h a t P i s i n v e r t i b l e means t h a t
# 0. Hence,
1
i s w e l l defined. Hence, t h e (numerical) f a c t o r s
[
c a n c e l i n (1) t o y i e l d 1PAP-ll = 1 ~ 1 .
and
Ip - I =
IPI
I P'9
A NOTE ABOUT LINEAR TRANSFORMATIONS
I n t h e previous u n i t we d i s c u s s e d how t h e same l i n e a r t r a n s formation could be coded by many d i f f e r e n t m a t r i c e s , depending
on t h e b a s i s being used t o d e s c r i b e t h e domain of t h e t r a n s formation.
W e showed t h a t i f A and B were any two such m a t r i c e s ,
.
-1
then t h e r e e x i s t e d a non-singular matrix P such t h a t B = PAP
I n f a c t , we used t h i s f a c t a s t h e motivation f o r our d e f i n i t i o n
of what i t meant f o r two m a t r i c e s t o be s i m i l a r .
During t h e d i s c u s s i o n we a l s o mentioned t h a t t h e transformation
i t s e l f d i d n o t depend on t h e b a s i s being used. Only t h e m a t r i x
depended on t h e choice of b a s i s .
Thus, one would l i k e t o f e e l
t h a t t h e r e should be some i n v a r i a n t about t h e matrix of t h e
t r a n s f o r m a t i o n ; t h a t i s , some f a c t t h a t would be t r u e f o r
every s i m i l a r matrix. What we have shown i n t h i s e x e r c i s e i s
t h a t t h e r e i s a t l e a s t one such i n v a r i a n t ( t h e r e a r e a l s o
Namely, s i n c e
o t h e r s b u t we s h a l l n o t pursue t h i s here)
.
1 PAP-' 1
=
IAI
we know t h a t i f A and B a r e s i m i l a r then t h e s e
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.3 ( L ) c o n t i n u e d
two m a t r i c e s have t h e same d e t e r m i n a n t .
I n o t h e r words, t h e
i n f i n i t e l y many d i f f e r e n t m a t r i c e s which code t h e same l i n e a r
t r a n s f o r m a t i o n a r e c h a r a c t e r i z e d (among o t h e r ways) by t h e f a c t
t h a t t h e y a l l have t h e same d e t e r m i n a n t .
3.5.4
(optional)
W e have
Hence,
d e t (A
+
all
+
I) = a2 1 Now
Solutions
Block 3: S e l e c t e d Topics i n Linear Algebra
Unit 5: Determinants
3.5.4
continued
(2) might n o t s e e m l i k e a very impressive r e s u l t , b u t
it has s e v e r a l important r a m i f i c a t i o n s . From our immediate p o i n t
In i t s e l f ,
of view, perhaps t h e most important consequence of ( 2 ) i s t h a t
it shows us t h a t " t h e determinant of a sum equals t h e sum of
t h e determinants" i s u s u a l l y a f a l s e statement. Indeed, we have
shown i n (2) t h a t
d e t (A
+
I ) = det(A)
+
det(1)
i s t r u e only when all + a 2 2 = 0.
diagonal elements of A i s z e r o ,
d e t (A
+
I)
That i s , u n l e s s t h e sum of t h e
# det(A) + d e t ( 1 ) .
of t h e diagonal
elements of a s q u a r e m a t r i x i s given a s p e c i a l name. I t i s
I t i s a l s o i n t e r e s t i n g t o n o t e t h a t t h e sum
c a l l e d t h e t r a c e o f t h e matrix.
We s h a l l e x h i b i t an i n t e r e s t i n g
property o f t h e t r a c e of a matrix i n p a r t (b).
Hence, w e conclude from (1) t h a t i f
then
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.4
continued
Therefore,
Thus f o r ( 2 ) w e conclude t h a t t h e t r a c e o f B-lAB,
, equals
T~ (B-'AB)
-3
+
written
1 5 = 12.
Note #1:
W e have w r i t t e n B-lAB whereas i n t h e p a r t w e have t a l k e d about
-1
t h e n B-lAB t a k e s t h e
BABml.
Notice t h a t i f w e l e t P = B
PAP-'.
I n o t h e r words, we may w r i t e e i t h e r BmlAB o r BAI3-l
o u r d e f i n i t i o n of s i m i l a r m a t r i c e s .
in
I n computing t h e m a t r i x
o f t h e t r a n s f o r m a t i o n , however, i t does make a d i f f e r e n c e
whether w e w r i t e B-lAE? o r BAB-'
b u t w e s h a l l say more about
t h i s i n the next unit.
Note #2:
I n what may s e e m t o be a c o i n c i d e n c e , t h e t r a c e of A i s a l s o
12 (namely, t h e d i a g o n a l elements of A a r e 4 and 8 ) . Thus, a t
l e a s t i n t h i s s p e c i a l c a s e , we have t h a t A and B-lAB have t h e
same t r a c e .
The i n t e r e s t i n g p o i n t i s t h a t t h i s i s n o t a c o i n c i -
dence. While w e s h a l l n o t b o t h e r p r o v i n g t h i s r e s u l t i n o u r
c o u r s e , t h e f a c t i s t h a t i f B i s any non-singular n by n
m a t r i x and A i s any n by n m a t r i x t h e n A and B - ~ A B always have
t h e same t r a c e .
Recalling o u r d e f i n i t i o n i n t h e previous u n i t t h a t A i s s i m i l a r
t o C means t h a t C = B - ~ A B , we s e e t h a t s i m i l a r m a t r i c e s have
t h e same t r a c e .
I n p a r t i c u l a r , t h e n , i n r e f e r e n c e t o o u r coding
a l i n e a r t r a n s f o r m a t i o n by a m a t r i x , t h i s shows t h a t t h e m a t r i x
which d e n o t e s t h e t r a n s f o r m a t i o n ,
while it v a r i e s with t h e
c h o i c e of b a s i s , has t h e same t r a c e , r e g a r d l e s s o f t h e b a s i s
Solutions
Block 3: S e l e c t e d Topics i n Linear Algebra
Unit 5: Determinants
3.5.4
continued
being used.
I n o t h e r words, when we t a l k about t h e t r a c e of t h e
matrix, it makes no d i f f e r e n c e which b a s i s i s being used.
In
s t i l l o t h e r words, t h e t r a c e of t h e matrix which d e s c r i b e s t h e
l i n e a r transformation i s a l s o an i n v a r i a n t with r e s p e c t t o t h e
basis.
a. Using t h e method of c o f a c t o r s and expanding along t h e f i r s t row,
we o b t a i n
W e may now expand
a
, @ ,@ ,
t h e f i r s t row t o o b t a i n :
and
@
by c o f a c t o r s along
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.5 (L) c o n t i n u e d
Using t h e r e s u l t s of ( 2 ) ,( 3 ) ,(41, and ( 5 ) i n (1) w e o b t a i n
Note #1
Our main aim i n h a v i n g you do p a r t ( a ) i s s o t h a t you can conv i n c e y o u r s e l f t h a t t h e method o f c o f a c t o r s , w h i l e s t r u c t u r a l l y
sound, i s extremely cumbersome t o apply i n even simple numerical*
cases.
T h a t i s , a 3 by 3 d e t e r m i n a n t can h a r d l y be c a l l e d
c o m p l i c a t e d , y e t t h e amount of computation i n p a r t ( a ) a l r e a d y
seems t o b e g e t t i n g q u i t e heavy.
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.5 (L) continued
Our p o i n t i s t h a t t h e method of c o f a c t o r s i s extremely important
when we a r e d e a l i n g with " l i t e r a l " m a t r i c e s , b u t i n t h e case
t h a t we a r e d e a l i n g with numerical examples, it i s much e a s i e r
t o "row-reduce" t h e determinant, j u s t a s we d i d with matrices.
I n t h i s r e s p e c t , we invoke t h e axiom t h a t t h e determinant i s
unchanged i f we r e p l a c e any row (column) by i t s e l f p l u s a
c o n s t a n t m u l t i p l e of any o t h e r row (column).
Here, again we
s e e t h e d i f f e r e n c e between m a t r i c e s and determinants; when we
row-reduce a m a t r i x we g e t d i f f e r e n t though e q u i v a l e n t ( i . e . ,
they span t h e same space) m a t r i c e s , b u t when we row-reduce a
determinant i n t h e above-mentioned c o n t e x t , w e do n o t change
t h e determinant.
With t h i s i d e a i n mind one u s u a l l y e v a l u a t e s
an n by n determinant by row reducing it t o a form i n which i n
one row ( o r column) a l l t h e e n t r i e s , b u t one, a r e zero.
I n p a r t (b) w e i l l u s t r a t e t h i s technique by g e t t i n g t h e d e t e r minant i n t o t h e form i n which a l l b u t t h e f i r s t e n t r y of t h e
f i r s t column a r e zero.
W e o b t a i n e d (ii)from (i)by r e p l a c i n g t h e 2nd row of
(i)by
t h e 2nd minus t h e 1st; t h e 3rd, by t h e 3rd minus t w i c e t h e
f i r s t ; and t h e 4 t h , by t h e 4th minus t h r e e t i m e s t h e f i r s t .
The f i r s t column of (7) has 0 ' s everywhere except f o r a 1
i n t h e f i r s t row.
Hence, i f w e expand (ii)by c o f a c t o r s
along t h e f i ~ s column
t
we obtain
1 1
1 1 2
o-[:
0
-4
-2
2
3
4
2
3
4
1
2
1 + 0 1
1
2
0
4
0
-4,
-2
-2
-01
1
1
2
'21 ' 8 1
l*
Solutions
Block 3 : S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.5 (L) c o n t i n u e d
Comparing ( 7 ) and ( 8 ) w e see t h a t t h e 4 by 4 d e t e r m i n a n t (i) i s
e q u a l t o t h e 3 by 3 d e t e r m i n a n t (iii)where
W e may a g a i n row-reduce
(iii)t o o b t a i n
and expanding ( i v ) by c o f a c t o r s a l o n g t h e f i r s t column, w e
obtain
Combining s t e p s ( 7 ) through (11) [and w e reproduce t h e s e
s t e p s s o t h a t you see t h e " b i g p i c t u r e " ] we o b t a i n
Solutions
Block 3: S e l e c t e d Topics i n Linear Algebra
Unit 5: Determinants
3.5.5 ( L ) continued
C l e a r l y (12) agrees with (6), b u t c e r t a i n l y t h e r e i s no doubt
t h a t t h e method of p a r t ( b ) i s more "sane" than t h a t of p a r t
(a)
.
Note #2:
A s a compromise between t h e techniques of ( a ) and ( b ) , we should
observe t h a t we may continue t h e row reduction technique u n t i l
w e have a diagonal 4 by 4 matrix. The i d e a i s t h a t f o r a
diagonal m a t r i x t h e determinant i s simply t h e product of t h e
d i a g o n a l elements. The key p o i n t h e r e i s t h a t i f we use t h i s
In other
method, no r e f e r e n c e a t a l l i s made t o c o f a c t o r s .
words, w e may e v a l u a t e t h e determinant without recourse t o
anything b u t t h e axioms under which t h e determinant f u n c t i o n
was derived.
More s p e c i f i c a l l y , w e have:
W e may now f a c t o r a .2 o u t of t h e bottom row of t h e l a s t d e t e r -
minant i n (13) t o o b t a i n
I
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.5 (L) c o n t i n u e d
Observations
1. The r e a s o n t h a t w e know t h a t t h e d e t e r m i n a n t of a d i a g o n a l
matrix is t h e product of t h e diagonal e n t r i e s is a s follows (we
u s e t h e 4 by 4 c a s e , b u t t h e g e n e r a l i z a t i o n s h o u l d b e c l e a r ) :
Given :
w e may " f a c t o r o u t " all from t h e f i r s t row; a22. from t h e
second row; a331 from t h e t h i r d row; and
from t h e f o u r t h
row t o o b t a i n
and t h i s .
2.
i n t u r n , e q u a l s alla22a33a44
since
111
= 1.
A c t u a l l y i f t h e e n t i r e s of A a r e a l l z e r o below t h e main d i a -
. ., ai
gonal (i e
= 0 i f i > j ) t h e n t h e d e t e r m i n a n t of A i s
s t i l l t h e p r o d u c t of t h e d i a g o n a l e n t r i e s .
Namely, given
w e may expand by c o f a c t o r s a l o n g t h e f i r s t column t o o b t a i n
successively
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
and f a c t o r i n g o u t 2 from t h e 4th row of (1) and -5 from t h e
5 t h row, w e o b t a i n
S i n c e (i)i s a t r i a n g u l a r m a t r i x (i.e.
, all
e n t r i e s below t h e
main d i a g o n a l a r e z e r o ) , i t s d e t e r m i n a n t i s t h e p r o d u c t of
d i a g o n a l e n t r i e s , i.e.,
(1)(1)(1)(-1)(-4)
r e s u l t w i t h (1) and ( 2 ) w e have t h a t
= 4.
Combining t h i s
I
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
(optional)
Our main aim i n t h i s e x e r c i s e i s t o t r y t o e x p l a i n how t h e
formula o f expanding by t h e use of c o f a c t o r s e v o l v e s from o u r
t h r e e b a s i c axioms used t o d e f i n e D.
is t h e l i n e a r property 0f.D.
that
The key s t e p , o f c o u r s e ,
For example, w e know by l i n e a r i t y
I f w e now w r i t e t h i s same r e s u l t i n t h e more c o n v e n t i o n a l
notion f o r determinants, we obtain that:
An i n t e r e s t i n g q u e s t i o n i s t h a t o f d e t e r m i n i n g how (1) could
b e d e r i v e d d i r e c t l y from o u r p r e v i o u s r e c i p e s w i t h o u t r e f e r e n c e
t o t h e D-notation.
T h i s i s answered i n p a r t ( a ) o f t h i s
exercise.
a.
Using a s h i n d s i g h t t h e f a c t s t h a t
"codes" D(allul,
rewrite
a 2 1 ~ 1 + a 2 2 ~ 2I ) and allUl
--
allUl
+ 0u2, w e
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
N o t i c e t h e p o s i t i o n of 0 i n t h e f i r s t row o f ( 3 ) .
column it f o l l o w s all,
In the f i r s t
b u t i n t h e second column, it p r e c e d e s
aI2.
i s t h e c o e f f i c i e n t of ul w h i l e
i s t h e c o e f f i c i e n t o f u2; s o t h a t w e may t h i n k of all i n
a
12
The r e a s o n f o r t h i s i s t h a t all
( 2 ) a s d e n o t i n g allul
+
Ou2; w h i l e a12 d e n o t e s Oul
+
a12u2.
Be
t h i s a s i t may, however, w e now u s e t h e r e s u l t d i s c u s s e d i n
e x e r c i s e 3.5.1 and rewrite ( 3 ) a s
W e may now ,rewrite t h e bottom row o f each d e t e r m i n a n t of
by r e p l a c i n g apl by a21
+
0 and a a 2 by 0
+
a22.
(4)
W e then obtain
and
I f w e now p u t t h e r e s u l t s of ( 2 ) through ( 6 ) t o g e t h e r w e have:
@
@
and
a r e 0. W e nay see t h i s by o b s e r v i n g
t h a t each h a s a column o f 0 ' s o r by o b s e r v i n g t h a t i n each
Determinants
d e t e r m i n a n t one row i s a s c a l a r m u l t i p l e o f t h e o t h e r ,
any e v e n t , w e are l e f t w i t h
In
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.7
continued
l
Determinant
05
h a s an i n t e r e s t i n g form.
of t h e i d e n t i t y determinant.
from @by
I t i s a permutation
That i s , we can o b t a i n t h e i d e n t i t y
interchanging t h e two rows.
Now, we know t h a t we
change t h e s i g n of t h e d e t e r m i n a n t whenever w e i n t e r c h a n g e two
rows.
Hence,
and s i n c e
w e conclude from ( 8 ) t h a t
b . I n a l l candor, it s h o u l d appear t h a t o u r d e r i v a t i o n of ( 9 ) was
l i k i n g s n a t c h i n g d e f e a t from t h e jaws of v i c t o r y !
After a l l ,
w e d e r i v e d t h i s r e s u l t i n t h e l e c t u r e . Indeed, t h e t e c h n i q u e
used i n t h i s e x e r c i s e was p r e c i s e l y t h e same a s t h a t used i n
t h e l e c t u r e e x c e p t f o r t h e symbolism used t o denote t h e d e t e r minant f u n c t i o n .
The i m p o r t a n t p o i n t i s t h a t i n t h e format used i n p a r t ( a ) of
t h i s e x e r c i s e w e b e g i n t o s e e t h e s t r u c t u r e which j u s t i f i e s
t h e s i g n convention of t h e method of c o f a c t o r s .
The key i d e a
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
i s t h a t whenever we i n t e r c h a n g e two rows (columns) o f a s q u a r e
m a t r i x w e change t h e s i g n o f t h e d e t e r m i n a n t of t h e m a t r i x . Thus,
g i v e n a p e r m u t a t i o n of t h e i d e n t i t y m a t r i x I , w e can count how
many t i m e s w e must i n t e r c h a n g e two rows t o o b t a i n t h e i d e n t i t y
matrix.
I f t h e number o f t i m e s i s even, t h e d e t e r m i n a n t i s 1
( s i n c e e v e r y i n t e r c h a n g e changes t h e s i g n ) w h i l e i f t h e number
of i n t e r c h a n g e s i s odd t h e d e t e r m i n a n t i s -1.
Before p u r s u i n g t h i s i n more d e t a i l , l e t us r e i n f o r c e t h e t e c h n i q u e by v e r i f y i n g t h e formula f o r computing W e have t h a t (10) i s e q u a l t o
which i n t u r n i s e q u a l t o
W e may n m r e w r i t e t h e second row o f each d e t e r m i n a n t i n (7)
as
i n which c a s e w e o b t a i n :
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7 c o n t i n u e d
S i n c e t h e f i r s t two rows o f
@
are s c a l a r m u l t i p l e s of one
a n o t h e r , w e have t h a t t h e v a l u e o f
0
i s z e r o (more f o r m a l l y ,
and a21 from t h e
w e f a c t o r o u t all from t h e f i r s t row o f
second row, so t h a t t h e d e t e r m i n a n t h a s t h e form:
9
i n which c a s e two rows o f t h e new d e t e r m i n a n t a r e e q u a l ) .
To compute
0 0 10
For example,
Now
and
11 w e rewrite t h e t h i r d row of each a s
or
0
11 becomes
@
a r e each z e r o .
Namely i n @ w e
see t h a t the
f i r s t and t h i r d rows a r e s c a l a r m u l t i p l e s of one a n o t h e r ( o r i n
0
t e r m s o f t h e t r a n s p o s e m a t r i x , t h e second column of
r e p r e s e n t s t h e z e r o v e c t o r ) ; w h i l e i n 1 4 t h e second and t h i r d
rows a r e s c a l a r m u l t i p l e s o f one a n o t h e r .
0 It is
all
@
which i s of i n t e r e s t t o us.
That i s , w e may f a c t o r
from t h e f i r s t row; a 2 3 from t h e second row, and a32 from
t h e t h i r d row t o o b t a i n :
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
Notice t h a t
@
i s a permutation of t h e i d e n t i t y .
In particular, may b e c o n v e r t e d i n t o simply by i n t e r c h a n g i n g t h e second and t h i r d rows.
More s p e c i - fically Combining t h e r e s u l t s of (13) , ( 1 4 1 , and (15) I w e See t h a t Note : W e c o u l d have s a i d t h a t and t h i s seems much s i m p l e r t h a n o u r method of d e r i v i n g ( 1 6 ) .
Keep i n mind, however, t h a t t h e purpose o f t h i s e x e r c i s e i s t o
j u s t i f y t h e method of expansion by c o f a c t o r s . I n o t h e r words,
t h e " s h o r t c u t " i s a form of c i r c u l a r r e a s o n i n g i n t h a t it
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.7
continued
presupposes t h e v a l i d i t y o f t h e method o f c o f a c t o r s .
I n any e v e n t w e can perform a s i m i l a r a n a l y s i s on d e t e r m i n a n t s
@
and
@
.
S p a r i n g you t h e d e t a i l s , t h e o v e r a l l t e c h n i q u e
t a k e s on t h e form:
[You may want t o o b s e r v e t h a t t h e r i g h t s i d e o f (17) i s e q u i v a l e n t
t o expanding t h e l e f t s c a l e by c o f a c t o r s w i t h r e s p e c t t o t h e
f i r s t row. ]
A s w e saw,
0
6
may b e r e w r i t t e n a s
r 1
all O
= O
a
a
O
22 a31 O 0
+
O
0
+
11
0
0
a
0
11
a 2 2 0 + 0
O
a
32
+
O
0
0
0
0
a 22 0
O
a23
Solutions
Block 3: S e l e c t e d Topics i n Linear Algebra
Unit 5: Determinants
3 . 5 . 7 continued
Similarly
0
7
becomes
= 0 s i n c e f i r s t two rows are s c a l a r multip l e s of each other. A
\
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
Both
@
a r e permutations of t h e i d e n t i t y matrix.
and
How can w e c o n v e r t
@
t o the identity.
Well, we would want
1
0
0 t o b e o u r f i r s t row, b u t i t i s now o u r second row.
Hence, w e would f i r s t i n t e r c h a n g e o u r f i r s t and second rows
t o obtain
On t h e o t h e r hand, t o c o n v e r t
, we
notice that 1
0
0
i s t h e t h i r d row, s o w e i n t e r c h a n g e i t w i t h t h e p r e s e n t f i r s t
row t o o b t a i n
Looking a t
@
w e would l i k e 0
1
0 t o b e t h e second row
r a t h e r t h a n t h e t h i r d , s o w e i n t e r c h a n g e t h e second and t h i r d
rows of
@
t o obtain
and combining (20) and (21) e s t a b l i s h e s t h a t
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.7
continued
Thus,
(19) becomes
'a12a21a33
+
a12a23a31'
F i n a l l y , l e a v i n g t h e d e t a i l s to. you,
0
8
reduces t o
A q u i c k " u n e t h i c a l " way o f v e r i f y i n g (23) i s t o expand
c o f a c t o r s a l o n g t h e f i r s t row.
@ by
Again, t h i s i s c i r c u l a r
r e a s o n i n g s i n c e o u r aim h e r e i s t o e s t a b l i s h t h e v a l i d i t y o f
t h e procedure.
~ f w e now combine (181, (221, and (231, w e o b t a i n t h e w e l l
known r e s u l t :
Note:
The major impact of what l o o k s l i k e a "shaggy dog" s t o r y h e r e
i n p a r t ( b ) i s t h a t t h e r e was a s y s t e m a t i c way o f p r e d i c t i n g
what (24) was going t o b e , and t h a t t h i s s y s t e m a t i c way e x t e n d s
a t once t o any n by n d e t e r m i n a n t . Using t h e 3 by 3 c a s e a s
an i l l u s t r a t i o n , we o b s e r v e t h a t t h e given d e t e r m i n a n t may b e
w r i t t e n a s t h e sum o f 27 v e r y s p e c i a l d e t e r m i n a n t s . Namely,
w e g e t one d e t e r m i n a n t f o r each way t h e r e i s of p i c k i n g
e x a c t l y one element from each row.
We t h e n f a c t o r o u t t h e
v a r i o u s c o e f f i c i e n t s and we f i n d t h a t each r e s u l t i n g d e t e r m i n a n t
i n v o l v e s a m a t r i x which e i t h e r h a s a t l e a s t one row (column) of
0 ' s o r else t h e m a t r i x i s a permutation of t h e i d e n t i t y m a t r i x .
The p e r m u t a t i o n o f t h e i d e n t i t y m a t r i x o c c u r s i f and o n l y i f
we had one o f t h e 27 s p e c i a l m a t r i c e s i n which t h e s e l e c t e d
e n t r i e s from each row a l s o were from d i s t i n c t columns.
t h i s means i s t h e following:
What
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7 c o n t i n u e d
I f w e look a t any t e r m which c o n s i s t s of t h r e e f a c t o r s o f t h e
aijls,
t h i s t e r m w i l l be a b s e n t from ( 2 4 ) u n l e s s b o t h t h e s e t of
f i r s t s u b s c r i p t s and second s u b s c r i p t s a r e p e r m u t a t i o n s o f t h e
numbers 1, 2, 3.
For example, n o t i c e t h a t t h e term alla12a23
does n o t a p p e a r i n (24) s i n c e t h e f i r s t s u b s c r i p t of t h e a ' s
form t h e s e t
{ 1 , 1 , 2 ) which i s n o t a p e r m u t a t i o n of { 1 , 2 , 3 )
.
R e c a l l t h a t a p e r m u t a t i o n o f a set S i s a 1-1, o n t o f u n c t i o n
f:S
-+
S , hence a t most a re-arrangement of t h e o r d e r o f { 1 , 2 , 3 )
Thus, w e see t h a t t h e r e w i l l be a s many d i s t i n c t t e r m s i n (24)
a s t h e r e a r e ways o f o b t a i n i n g a permutation of t h e i d e n t i t y
m a t r i x . To t h i s end, n o t i c e t h a t i n t h e 3 by 3 c a s e t h e r e
e x a c t l y 3! = 6 such ways.
Namely, t h e row (1 0
0) may be
I
p l a c e d i n any one o f t h e t h r e e rows; once t h i s i s done, t h e row
(0
1 0) may b e p l a c e d i n any one of t h e two remaining rows;
and once t h i s i s done (0
position.
0
1) must b e p l a c e d i n t h e l a s t
N o t i c e t h a t (24) b e a r s o u t this remark s i n c e t h e
d e t e r m i n a n t i n (24) i n v o l v e s s i x t e r m s .
A l l t h a t remains t o account f o r i n (24) i s how t o determine
whether one o f t h e t e r m s o c c u r s w i t h t h e p l u s s i g n o r w i t h t h e
negative sign.
Conceptually t h i s i s easy t o do.
W e look a t
t h e p e r m u t a t i o n d e t e r m i n a n t from which t h i s t e r m r e s u l t s and
t h e n s e e how many i n t e r c h a n g e s a r e n e c e s s a r y t o c o n v e r t it t o
t h e i d e n t i t y . I f it t a k e s an odd number then t h e s i g n i s
n e g a t i v e ( s i n c e each i n t e r c h a n g e changes t h e s i g n of t h e d e t e r minant) w h i l e i f it t a k e s an even number t h e s i g n i s p o s i t i v e .
( I t i s an i n t e r e s t i n g a s i d e t h a t t h e number o f i n t e r c h a n g e s
depends on how w e e l e c t t o make t h e changes, b u t whether t h e
number o f i n t e r c h a n g e s i s even o r odd does n o t depend on how
w e choose t o make t h e i n t e r c h a n g e s . )
The q u e s t i o n i s how can we t e l l t h e s i g n w i t h o u t having t o
reproduce t h e m a t r i x .
T h i s , t o o , i s n o t very d i f f i c u l t
-
once
you g e t t h e c e n t r a l i d e a .
A l l we do i s o r d e r t h e f a c t o r s s o t h a t e i t h e r t h e f i r s t sub-
s c r i p t s o r t h e second s u b s c r i p t s o c c u r i n t h e " n a t u r a l " o r d e r ;
e
l 1 2, 3
We t h e n look a t t h e o t h e r s u b s c r i p t and s e e
how many i n t e r c h a n g e s a r e n e c e s s a r y t o g e t them i n t o t h e n a t u r a l
order.
One r e l a t i v e l y q u i c k way o f doing t h i s i s t o c o u n t t h e
.
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.7
continued
number of i n v e r s i o n s (which means t o count t h e number o f t i m e s
a lesser number comes a f t e r a g r e a t e r number).
S i n c e t h e i d e a o f i n v e r s i o n s may seem new t o you, l e t u s pause
t o i l l u s t r a t e t h i s i d e a w i t h a s p e c i f i c example. L e t us c o n s i d e r
t h e p e r m u t a t i o n on t h e s e t
1,2,3,4,5
which l e a d s t o t h e re-
arrangement
W e notice that:
1.
S t a r t i n g w i t h o u r f i r s t member 4 , t h r e e lesser numbers
f o l l o w it (namely 2, 1, and 3)
.
So t h i s g i v e s us 3 i n v e r s i o n s
so far.
2.
Our second number 5 i s a l s o followed by t h r e e lesser ones.
So t h i s g i v e s u s t h r e e more i n v e r s i o n s , f o r a t o t a l of s i x ,
so far.
3.
Our n e x t e n t r y 2 i s followed by o n l y one l e s s e r number
(namely 1 ) . T h i s y i e l d s one more i n v e r s i o n , f o r a t o t a l o f
7.
4.
Our n e x t number 1 i s followed by n o lesser numbers.
(This
i s t h e n i c e t h i n g about 1. I t c a n ' t b e followed by a lesser
number.
I n a s i m i l a r way i f n i s t h e g r e a t e s t number i n o u r
s e t , e v e r y t h i n g which f o l l o w s it i s a l e s s e r number.) So o u r
t o t a l number o f i n v e r s i o n s i s s t i l l 7.
5. Our l a s t number i s 3 and s i n c e it i s t h e l a s t number l i s t e d ,
n o t h i n g f o l l o w s it. I n p a r t i c u l a r , no l e s s e r number f o l l o w s it.
Thus, o u r t o t a l number o f i n v e r s i o n s i s 7 which i s odd.
Obviously we could have re-arranged (25) i n t o t h e n a t u r a l
o r d e r i n a less m y s t i c a l way. For example, we might have
n o t i c e d t h a t w e wanted 5 t o come l a s t b u t s i n c e it was l i s t e d
second i n ( 2 5 ) , w e would have t o i n t e r c h a n g e t h e second and
f i f t h e n t r i e s o f (25)
.
That i s , o u r f i r s t i n t e r c h a n g e would
yield
Looking a t (26) w e know t h a t 4 s h o u l d occupy t h e f o u r t h
p o s i t i o n , b u t it i s p r e s e n t l y i n t h e f i r s t p o s i t i o n , s o w e
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
r e a l i z e t h a t w e must i n t e r c h a n g e t h e f i r s t and f o u r t h terms o f
(26) , and t h i s y i e l d s
W e t h e n o b s e r v e t h a t 2 and 3 s h o u l d b e i n t e r c h a n g e d i n (27) f o r
us t o obtain t h e n a t u r a l order
s o t h a t a l t o g e t h e r w e have had t o make t h r e e i n t e r c h a n g e s t o
g e t (25) i n t o t h e n a t u r a l o r d e r .
T h i s checks w i t h t h e f a c t
t h a t w e had an odd ( 7 ) number o f i n v e r s i o n s .
Keep i n mind t h a t w e c o u l d have p u t (25) i n t o n a t u r a l o r d e r
i n o t h e r s y s t e m a t i c ways.
For example w e might have decided
t h a t w e f i r s t wanted 1 t o appear f i r s t , i n which c a s e we would
have i n t e r c h a n g e d t h e 1 and 4 t o o b t a i n
W e might t h e n have i n t e r c h a n g e d 5 and 2 t o g e t 2 i n t h e r i g h t
place, thus yielding
W e might t h e n have i n t e r c h a n g e d 3 and 5 t o p u t 3 i n t h e r i g h t
p l a c e , and t h i s would have y i e l d e d
s o t h a t a g a i n t h r e e i n t e r c h a n g e s b r i n g s about t h e n a t u r a l o r d e r .
I n f a c t , t h e method of i n v e r s i o n i s a s p e c i a l way of c o u n t i n g
interchanges.
Namely, t o g e t 5 i n t o i t s p r o p e r p l a c e , w e must
have it "jump over" e v e r y l e s s e r number and t h i s i s e q u i v a l e n t
t o i n t e r c h a n g i n g it w i t h each l e s s e r number.
That i s t h e number
o f i n v e r s i o n s ( 7 ) corresponds t o t h e f o l l o w i n g s u c c e s s i o n of
interchanges :
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
I n t e r m s o f o u r d e t e r m i n a n t i d e a , i f we were expanding t h e 5 by
5 determinant l a
ij
I
t h e n t h e s i g n of
would b e n e g a t i v e because w i t h t h e f i r s t s u b s c r i p t s i n t h e
n a t u r a l o r d e r t h e second ones a r e t h e permutation 4
5
2
1
3
which i s odd.
I n more g e n e r a l terms, t h e expansion of an n by n m a t r i x w i l l
c o n s i s t of n! t e r m s s i n c e t h i s i s t h e number o f p e r m u t a t i o n s
o f In.
Half o f t h e s e n! t e r m s w i l l b e p o s i t i v e and t h e o t h e r
half negative.
To see which s i g n a term h a s , w e l i n e up t h e
f a c t o r s i n t h e o r d e r i n which, s a y , t h e f i r s t s u b s c r i p t s form
...,
n ( i f t h i s cannot b e done, t h e t e r m i s
t h e n a t u r a l o r d e r 1,
n o t one of t h e n! b e i n g e v a l u a t e d * ) . We t h e n look a t t h e p e r m u t a t i o n formed by t h e second s u b s c r i p t s and count t h e number
of inversions.
I f t h e number of i n v e r s i o n s i s even, w e u s e
the plus sign, otherwise, the negative sign.
*
I n t h e method o f p a r t ( b ) , g e n e r a l i z e d t o t h e n by n c a s e ,
we o b s e r v e t h a t we g e t o n e s p e c i a l d e t e r m i n a n t f o r e a c h way
t h a t we c a n f o r m a m a t r i x u s i n g o n e e n t r y f r o m e a c h row.
Since
t h e r e a r e n rows a n d n e n t r i e s f r o m e a c h row, t h e r e a r e n n s u c h
s p e c i a l determinants (note t h a t t h i s accounts f o r t h e 27 i n our
O f t h e s e nn s p e c i a l d e t e r m i n a n t s , a l l
example w h e r e i n n = 3 ) .
w i l l e q u a l 0 e x c e p t f o r t h e n! w h i c h a r e p e r m u t a t i o n s o f
n
11 I
Solutions
Block 3: S e l e c t e d T o p i c s i n L i n e a r Algebra
U n i t 5: D e t e r m i n a n t s
3.5.7
c.
continued
Given
w e may t h i n k o f r c o d i n g t h e row (1 0 0 0 0 ) ; r2, t h e row
1 ( 0 1 0 0 O ) , etc. I n t h i s way, I i s r e p r e s e n t e d by t h e s e q u e n c e o f rows
Our d e t e r m i n a n t h a s t h e rows o f I i n t h e o r d e r
A s a s h o r t c u t , w e may u s e j u s t t h e s u b s c r i p t s i n (28) and (29),
and w e see t h a t w e must o n l y c o u n t t h e i n v e r s i o n i n
which i s
( i - e . , 2 and 1 f o l l o w s 3; t h e y f o l l o w 4 ; t h e y f o l l o w 5; and 1
follows 2 ) .
Hence, t h e g i v e n d e t e r m i n a n t i s -1.
More c o n c r e t e l y :
Solutions
Block 3: S e l e c t e d T o p i c s i n L i n e a r Algebra
U n i t 5: Determinants
3.5.7
continued
Summary :
What w e have t r i e d t o g i v e you a f e e l i n g f o r i n t h i s e x e r c i s e i s
where t h e formula f o r e v a l u a t i n g d e t e r m i n a n t s comes from, and
how t h e method of expansion by c o f a c t o r s i s a c o n v e n i e n t way
o f keeping t r a c k o f t h e s e r e s u l t s w i t h o u t c o n s c i o u s l y having
t o d e t e r m i n e whether a g i v e n permutation i n v o l v e s an odd o r an
even number o f i n t e r c h a n g e s t o c o n v e r t it t o n a t u r a l o r d e r .
Keep i n mind, t h e f a c t t h a t we've s t r e s s e d b e f o r e .
Namely, w e
worry a b o u t c o f a c t o r s , e t c . , u s u a l l y when w e a r e working w i t h
l i t e r a l expressions.
I n t h e c a s e of c.oncrete numbers, w e o f t e n
u s e row-reduction methods, e t c . s i n c e t h i s o f t e n y i e l d s t h e
c o r r e c t answer w i t h much less l a b o r .
3.5.8
(optional)
I n o u r s t u d y o f i n v e s t i g a t i n g t h e l i n e a r independence o f t h e
set
we had t o e v a l u a t e t h e Wronskian d e t e r m i n a n t
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.8
continued
W e c o u l d t h e n f a c t o r e rlx
from t h e f i r s t column of (1), er2X
from
t h e second column, etc. t o o b t a i n
The d e t e r m i n a n t i n ( 2 ) i s c a l l e d t h e Vandemonde d e t e r m i n a n t and
i t s v a l u e c o n s i s t s o f t h e p r o d u c t of a l l t e r m s o f t h e form ri
where i > j.
-
For example,
etc. T h i s r e s u l t shows t h a t t h e Wronskian of f e
'lX
e
r x
n
equals r
j
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.8
continued
= r
s i n c e i # j. I n o t h e r words, i f ri
i
j
t h e n t h e d e t e r m i n a n t i n (1) i s unequal t o 0.
0
++
r
# rj
f o r i # j.
Of c o u r s e , one does n o t need t h e s t u d y o f l i n e a r independence t o
m o t i v a t e t h e Vandemonde d e t e r m i n a n t .
This determinant i s
v a l u a b l e i n i t s own r i g h t and t h e purpose o f t h i s e x e r c i s e i s
t o v e r i f y i t s v a l u e , a t l e a s t i n t h e c a s e s ( 2 ) and ( 3 ) mentioned
above [Case (1) i s , h o p e f u l l y , completely t r i v i a l ]
.
a. S i n c e w e a r e more used t o row-reduction t h a n t o column r e d u c t i o n
w e may u s e t h e f a c t t h a t a m a t r i x and i t s t r a n s p o s e have t h e
same d e t e r m i n a n t ( o t h e r w i s e w e do u s e column r e d u c t i o n , which
i s e q u i v a l e n t t o t h e row-reduction o f t h e t r a n s p o s e ) . I n any
event,
W e may now f a c t o r r2
r3
-
-
r from t h e second row o f (3) and
1 r from t h e t h i r d row t o o b t a i n
1
( s i n c e t h e two rows a r e e
q
u
a
l
) = 0
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3.5.8
=
continued
(r2- rl) (r3- rl)
{
(r3- r2) 1
= (r3- rl) (r3- r2)(r2- rl)
r - r
2
1
r
rl
3
r - r
4
1
-
1 r2+r
= ( r -
r
r
- r
r
- r
1
r 2+ r 2r 1+ rl
2
1 r3+ r 1 r32+r 3r 1+ rl
1
r4+ rl
r4'+
2
2
r4r1+ rl 2
.
S i n c e t h e d e t e r m i n a n t i n ( 4 ) h a s a second column, each o f whose
e n t r i e s i s t h e sum o f two numbers, w e may w r i t e t h e d e t e r m i n a n t
a s t h e sum of two s i m p l e r d e t e r m i n a n t s , one of which i s z e r o .
Namely,
Solutions
Block 3: S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3 . 5 . 8 continued
= 0 ( s i n c e t h e second column i s a m u l t i p l e o f t h e
first)
Now,
Hence,
( 5 ) may b e r e w r i t t e n a s
Solutions
Block 3 : S e l e c t e d Topics i n L i n e a r Algebra
Unit 5: Determinants
3 . 5 . 8 continued
Again, t h e d e t e r m i n a n t i n ( 6 ) h a s a column i n which each e n t r y
i s t h e sum o f t h r e e numbers.
Hence, w e may rewrite i t a s t h e
sum o f t h r e e s i m p l e r d e t e r m i n a n t s .
.
Namely,
-
f
These a r e b o t h 0 s i n c e i n each two rows a r e t h e s&e.
Combining t h e r e s u l t s o f (5), ( 6 ) , and ( 7 ) with' ( 5 ) w e o b t a i n
the desired result.
Perhaps t h e most s i g n i f i c a n t p a r t of t h i s e x e r c i s e ( a s i d e from
t h e a c t u a l r e s u l t ) i s t o s e e how one can avoid much p a i n f u l comp u t a t i o n by h a v i n g t h e p a t i e n c e t o invoke t h e v a r i o u s theorems
r a t h e r t h a n t o t r y t o "bludgeon o u t " t h e answer from t h e b a s i c
d e f i n i t i o n of a d e t e r m i n a n t o r from t h e method of c o f a c t o r s .
You might want t o t r y p a r t (b) by working d i r e c t l y from t h e
method of c o f a c t o r s w i t h o u t row-reducing o r f a c t o r i n g .
I f you
do, you w i l l soon n o t i c e how i n v o l v e d t h e a r i t h m e t i c g e t s .
Moreover, i n t h e c a s e o f an n by n d e t e r m i n a n t i n which n
exceeds 4 , t h e a r i t h m e t i c blows o u t of p r o p o r t i o n very r a p i d l y
i f w e ' r e not careful.
The t e c h n i q u e used by us i n o u r s o l u t i o n
of ( a ) and ( b ) g e n e r a l i z e s very n i c e l y t o h i g h e r dimensions,
and - a t l e a s t r e l a t i v e t o most o t h e r methods
keep t h e a r i t h m e t i c f a i r l y s i m p l e .
-
it manages t o
I n summary, i f you have t o work much w i t h numerical d e t e r m i n a n t s ,
i t i s w i s e t o c u l t i v a t e t e c h n i q u e s f o r row r e d u c t i o n , f a c t o r i n g ,
and r e w r i t i n g d e t e r m i n a n t s a s a sum o f s i m p l e r d e t e r m i n a n t s ,
p r e f e r r a b l y a s a sum i n which most of t h e t e r m s a r e zero!
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Complex Variables, Differential Equations, and Linear Algebra
Prof. Herbert Gross
The following may not correspond to a particular course on MIT OpenCourseWare, but has been
provided by the author as an individual learning resource.
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.