adjoint matrix nnijaA

nnnn

n

n

AAA

AAA

AAA

A

21

22212

12111 . ofcoplement

algebraic is

ij

ij

a

A

adjoint matrix

what?toattention

pay should we, ing when writSo A

Yes! The array of algebraiccomplements!

example: Write the adjoint matrix of a 2*2 matrix.

dc

baA

ac

bdA

AA

nnnn

n

n

nnnn

n

n

AAA

AAA

AAA

aaa

aaa

aaa

21

22212

12111

21

22221

11211

A

A

EA AA

EAAAAA

An important formula!

)(

1111

tdeterminan eVandermond.example

1

113

12

11

223

22

21

321

jniji

nn

nnn

n

n

n

aa

aaaa

aaaa

aaaa

D

• 1.The elements are arranged as ascending order, and exponentials are arithmetical series.

• 2.The result may be positive,negative or zero.• 3. A product of n(n-1)/2 items.

Our task is to calculate other determinants by Vedermonde determinant. So we should memorize the form and result of Vedermonde determinant.Can you identify Vedermonde determinant？Can you solve problems by the result of Vedermonde determinant ？

Such as：

642718

16914

4312

1111

D

)34)(14)(13)(24)(23)(21( 12

1111

4321

)4()3()2()1(

)4()3()2()1(2222

3333

aaaa

aaaa

aaaa

D

?

641641

27931

1111

8421

D 12

3333

2222

)4()3()2()1(

)4()3()2()1(

4321

1111

aaaa

aaaa

aaaaD

3333

2222

)1()2()3()4(

)1()2()3()4(

1234

1111

aaaa

aaaa

aaaa

!1!2!3 12

Cramer principleConsider the following linear system:

,

,

,

2211

22222121

11212111

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

Similar to binary system, the nth element equation can be expressed by determinant.

Theorem 1. If the coefficient determinant

,0

21

22221

11211

nnnn

n

n

aaa

aaa

aaa

D

then the system has unique solution:

.,,, 22

11 D

Dx

D

Dx

D

Dx n

n

where

nnjnnjnn

njj

j

aabaa

aabaa

D

)1()1(1

1)1(11)1(111

There are two propositions to prove. One is the existence of solution, the other is the uniqueness. And the solution can be written as

D

Dx jj ).,,2,1( nj

To proveD

Dx j

j

are solutions, we should prove the equations

11

2

12

1

11 bD

Da

D

Da

D

Da n

n

From the equation above, we can get

.012121111 nnDaDaDaDb

Therefore we construct a (n+1)th order determinant

),,2,1( nj

nnnnn

n

n

n

aaab

aaab

aaab

D

21

112111

112111

1

This determinant equals zero. Computeit by the 1st row, we obtain

10 nD

nnnn

n

n

aaa

aaa

aaa

b

21

22221

11211

1

nnnn

n

n

aab

aab

aab

a

2

2222

1121

11

nnnn

n

n

aab

aab

aab

a

1

2212

1111

12

1,1

1,2212

1,1111

12)1(

nnnn

n

n

nn

aab

aab

aab

a

Then prove the uniqueness of solution, and

,D

Dx jj .jj DxD

By ,0D the first proposition is proved.

that is

nnDaDaDaDb 12121111 0

nnjnjnjjnn

njjjj

j

aaxaaa

aaxaaa

xD

)1()1(1

1)1(11)1(111

.jD

,11 nknjkjkk xaxaxab

).,,2,1( nk

Theorem 2. If the coefficient determinantis nonzero, then the system has only one solution.

0

,0

,0

2211

2222121

1212111

nnnnn

nn

nn

xaxaxa

xaxaxa

xaxaxa

System:

is called homogeneous linear system.

Theorem 3. If the coefficient determinantof a homogeneous linear equation ,0Dthen the system has one unique solution.

solution? nonzero has system the

if, be shouldWhat 1example ：

,02

,0

,0

zyx

zyx

zyx

12

11

11

D 13 0 .1

So when ,1the system has nonzero solution.

example 2： Prove that the following system has zero solution only.

Solve: If the system has nonzero solution, the coefficient determinant must be zero.

,0)2

1(

,0)2

1(

,0)2

1(

2211

2222121

1212111

nnnnn

nn

nn

xaxaxa

xaxaxa

xaxaxa

integers) all are ( ija

tdeterminant coefficien theis Here :Prove

2

1

2

12

1

21

22221

11211

nnnn

n

n

aaa

aaa

aaa

D

1222

2122

2212

2

1

21

22221

11211

nnnn

n

n

n

aaa

aaa

aaa

0

So the system has zero solution only.

Practice determinant：

25

25

35

35

23

23

32

32

2

nD

1.

The (n+1)th column plus to the nth,the 2nth column plus to the 1st one.

10

10

35

35

row,1st theminus row

th )1( the

n

nn5)1(

The question can also be solved by computing it according to the row or column.

Compute the determinant by the 1st row.

23

23

23

32

32

32

2

nD

1220

23

23

32

32

32

2

n

1203

23

23

32

32

32

3

n

224 nD 229 nD 225 nD

422)5( nD 62

3)5( nD

21)5( Dn

n)5(

5

23

322

D

72

572

572

572

57

.2

nD

17 nD

72

5

72

57

52

5

21 107 nn DD

(compute it by the 1st row)

21 107 nnn DDD

)5(25 211 nnnn DDDD

)5(2 322

nn DD )5(2 122 DDn n2

3972

572 D 71 D

45 12 DD

)5(52 211 nnnn DDDD

)2(5 322

nn DD )2(5 122 DDn n5

252 12 DD

15 nn DD n2

12 nn DD n5 3

25 11

nn

nD

Keep the method in heart!