adjoint matrix yes! the array of algebraic complements!
TRANSCRIPT
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!