Chapter 2Determinant

Objective: To introduce the notion of determinant, and some of its properties as well as applications.

Test for singularity of a matrix instead of by definition.

Find the area of a parallelogram generated by two vectors.

Find the volume of a parallelopipe spanned by three vectors.

Solve Ax=b by Cramer’s rule.

Several applications of determinant

Introduction to Determinant (to determine the singularity of a matrix)

Consider .

If we define det(A)=a,

then A is nonsingular.

11)( FaA

det 0.A

Let

Suppose ,

then A

If we define

then A is nonsingular

22

2221

1211

F

aa

aaA

011 a

)()( 1122112 aEaE

21122211

1211

0 aaaa

aa

.)det( 21121211 aaaaA

.0)det( A

Case2 2×2 Matrices

Suppose but ,

then A and

Thus A is nonsingular Suppose A is singular & det(A)=0.

To summarize, A is nonsingular

12

2221

0 a

aa12E .)det( 2112aaA

011 a 012 a

.0)det( A

02111 aa

0)det( A

Case2 2×2 Matrices (cont.)

Let

-Suppose , A

33

333231

232221

131211

F

aaa

aaa

aaa

A

011 a

)()(11

21

11

31

1213 aa

aa EE

11

13313311

11

12313213

11

13212311

11

21122211

0

0131211

aaaaa

aaaaa

aaaaa

aaaaa

aaa

Case3 3×3 Matrices

From 2x2 case,

A I

Then A is nonsingular

Case3 3×3 Matrices (cont.)

0det11

13313311

11

12313213

11

13212311

11

21122211

aaaaa

aaaaa

aaaaa

aaaaa

.0)det( A

11 22 33 11 32 23 12 21 33

12 31 23 13 21 32 13 31 22 det( ) 0

a a a a a a a a a

a a a a a a a a a A

define

that A I

0)(

0,0)(

0,0)(

312111

312111

2111

aaaiii

aaaii

aai

.0)det( A

Easily Shown for Cases

For ,

where

22FA

21122211det aaaaA

)det()1()det()det( 1

2

11

112121111 j

jj

j MaMaMa

)(&)( 21122211 aMaM

Recall

For ,

where , ,

33FA

)()(det 23313321122332332211 aaaaaaaaaaA

)( 2231322113 aaaaa )det()det()det( 131312121111 MaMaMa

)det()1( 11

11

j

n

jj

j Ma

3332

232211 aa

aaM

3331

232112 aa

aaM

3231

222113 aa

aaM

Recall (cont.)

Definition: Let , , and let

the matrix obtained from A by

deleting the row & column containing The is called the minor of The cofactor of is denoted as

nnFA

.ija

)1()1( nnij FM

ijAija

)det()1( ijji

ij MA

Generalization

3n

det( )ijM

ija

Definition: The determinant of is defined as

, if n=1

, if n>1

Note: det is a function from to .

nnFA

Adetj

n

jjAa

a

11

1

11

nnF F

Theroem2.1.1:Let ,

Hint: By induction or sign-type definition.

2 , nFA nn

1 1

det( )

for 1,2,... . and 1,2,... .

n n

ik ik kj kjk k

A a A a A

i n j n

Theroem2.1.2: Let ,and

Pf: By induction,

n=1,ok!

Suppose the theorem is true for n=k.

If n=k+1,

By induction

The result then follows.

nnFA )det()det( AAT

1

111

11

1

11

1 )det()1()det()1()det(k

j

Tjj

jj

k

jj

j MaMaA

)det( TA

Theroem2.1.3: Let be a triangular matrix.

Then

Hint:expansion for lst row or column and induction on n.

Theroem2.1.4: (i)If A has a row or column consisting

entirely of zeros, then

(ii)If A has two identical rows or columns,

then

Hint for (ii): By mathematical induction.

nnFA .)det(

1

n

iiiaA

.0)det( A

.0)det( A

Note that

For example, ,

Question: Is

BABA detdet)det(

00

01A

10

00B

?)det()det()det( BAAB

§2-2 Properties of Determinants

Lemma2.2.1:

Let , then

nnFA

jninjijk

n

kik AaAaAa

111

ji if ,

ji if ,0 )det(A

Pf: Case for i=j follows directly from the definition

of determinant.

For ,

define to be the matrix obtained

from A by replacing the jth row of A by ith row

of A. (Then has two identical rows)

expansion along jth row

ji A

jninji AaAaA 11)det(0

jninji AaAa 11

Proof of Lemma2.2.1

A

Proof of Lemma2.2.1(cont.)

11 12 1

1 2

1 2

1 2

n

i i in

i i in

n n nn

a a a

a a a

A

a a a

a a a

jth row

Note that by Th. 2.1.3

by Th. 2.1.3

先對非交換列展開 數學歸納法

)det())(det( IEi

1))(det( ijE

1)det( ijE

)det())(det(1

AAaAE ij

n

jiji

)det())(det( AEi

jkjk

n

kikij AaaAE )())(det(

1

jkjkjkik AaAa

)det())(det()det( AEA ij 0

Lemma 2.2.1

:),(

:),(

det)det(

ia

ja

AEij

)det()det()det(

:),(

:),(

det AEA

ja

ia

ij

)1()1()1( jiijji EEE

rowith

rowjth

rowjth

rowith

Thus, we have

If E is an elementary matrix

In fact, det(AE)=det(A)det(E)

Question:

)det()det()det( AEEA

?)det()det()det( BAAB

Theorem2.2.2: is singular

Pf:Transform A to its row echelor from as

If A is singular

If A is nonsingular

The result then follows.

nnFA .0)det( A

AEEU k 1

1

det( ) ( det( ))det( )k

ii

U E A

0)det(0)det(

.

AU

rowzeroacontainsU

0)det(

1)det(

.1 of

A

U

isUelementsdiagonaltheall

0

Theorem2.2.3:Let .Then

Pf: If B is singular

AB is singular

If B is nonsingular

)det()det(0)det( BAAB

)det()det(

)det()det(

)det()det(

)det()det( 1

BA

EA

EA

EAEAB

i

i

k

nnFBA ,

)det()det()det( BAAB

.2.2.2Th

.,1 matriceselementaryareEwhereEEB ik

Objective:

Use determinant to compute and

solve Ax=b.

1A

§2-3 Cramer’s Rule

Def: Let .The adjoint of A is defined to be

where are cofactor of

nnFA

ijA .ija

11 21 1

12 22 2

1 2

( )

n

n

n n nn

A A A

A A Aadj A

A A A

The Adjoint of a Matrix

By Lemma2.2.1, we have

jk

n

kikAa

1

jiif

0

)det(A

jiif

If A is nonsingular, det(A) is a nonzero scalar

( ( )) det( )A adj A A I

1det( )

1 1det( )

( ( ))

( )

A

A

A adj A I

A adj A

For a 2×2 matrix :

If A is nonsingular , then

Example 1 (P.116)

22 12

21 11

( )a a

adj Aa a

22 121 1det( )

21 11A

a aA

a a

det( ) 0A

Q: Let , compute adj A and A-1.

Sol:

321

223

212

A

Example 2 (P.116)

1 1 1det( ) 5

2 1 2

( ) 7 4 2

4 3 1

2 1 2

( ) 7 4 2

4 3 1A

adj A

A adj A

Theorem2.3.1:(Cramer’s Rule)

Let be nonsingular and . Denote the

matrix obtained by replacing the ith column of A by .Then

the unique sol. of is

Pf:

nnFA nb F

iA

nix AA

ii ,,1for ,)det()det(

Ax b

1 1det( )

1det( )

1

1det( )

1

det( )det( )

( )

A

n

i j ijAj

n

j jiAj

iAA

x A b adj A b

x b adjA

b A

b

Q: Use Cramer’s rule to Solve

932

622

52

321

321

321

xxx

xxx

xxx

Example 3 (P.117)

Sol:

2,1,1

8)det(

4)det(

4

329

126

125

det)det(

4)det(

)det()det(

3)det()det(

2)det()det(

1

3

2

1

321

AA

AA

AA xxx

A

A

A

A

Example 3 (cont.)

Let .Then volume of the parallelopipe

spanned by and is

Let .Then the area of the parallelogram

spanned by and is

3,, cba

)det()( cbacbaV

2, ba

a

b

)det( baA

ba

, c

For example, the message

Send Money

might be coded as

5, 8, 10, 21, 7, 2, 10, 8, 3

here the S is represented by a “5”, the E is represented

by a “8”, and so on.

Application 1: Coded Message (P.118)

Application 1: Coded Message (cont.)

If A is a matrix whose entries are all integers and

whose determinants is ± 1, then, since ,

the entries of A-1 will be integers.

Let

1 A adj A

1 2 1 5 21 10

2 5 3 and B= 8 7 8

2 3 2 10 2 3

A

Application 1: Coded Message (cont.)

We can decode it by multiplying by A-1

We can construct A by applying a sequence of row operations on identity matrix.

Note:

1 1 1 31 37 29 5 21 10

2 0 1 80 83 69 8 7 8

4 1 1 54 67 50 10 2 3

det( ) det( ) 1A I

A-1 AB(encoding Message)