Matrices CHAPTER 8.1 ~ 8.8. Ch8.1-8.8_2 Contents 8.1 Matrix Algebra 8.1 Matrix Algebra 8.2 Systems...
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Transcript of Matrices CHAPTER 8.1 ~ 8.8. Ch8.1-8.8_2 Contents 8.1 Matrix Algebra 8.1 Matrix Algebra 8.2 Systems...
Matrices
CHAPTER 8.1 ~ 8.8
Ch8.1-8.8_2
Contents
8.1 Matrix Algebra8.2 Systems of Linear Algebra Equations8.3 Rank of a Matrix8.4 Determinants8.5 Properties of Determinants8.6 Inverse of a Matrix8.7 Cramer’s Rule8.8 The Eigenvalue Problem
Ch8.1-8.8_3
8.1 Matrix Algebra
MatrixThe forms
(x1 x2 … xn) or(1)
Each array in (1) is called a matrix.
nx
x
x
2
1
Ch8.1-8.8_4
Entry or Element: amn, the members in the arraySize: m nSquare matrix: n n, n is called the order.Main diagonal entries: ann
A matrix is any rectangular array of numbers or
functions
(2)
DEFINITION 8.1Matrix
mnmm
n
n
aaa
aaa
aaa
21
22221
11211
Ch8.1-8.8_5
A n 1 matrix is called a column vector.
A 1 n matrix is called a row vector.
DEFINITION 8.2Column and Row Vectors
na
a
a
2
1
)( 21 naaa
Two matrices A, B are equal if aij = bij .
DEFINITION 8.4Equality of Matrices
Ch8.1-8.8_6
Example 1
(a) The following matrix are not equal, since they are not of the same size.
(b) The following matrix are not equal, since the corresponding entries are not all equal.
11
11
111
111
43
21
34
21
Ch8.1-8.8_7
The sum of two mn matrices A, B is
A + B = (aij + bij)m n
DEFINITION 8.4Matrix Addition
Ch8.1-8.8_8
Example 2
(a)
211
539
874
,
5106
640
312
BA
395
1179
566
25)1(1016
563490
)8(37142
BA
Ch8.1-8.8_9
Example 2 (2)
(b)
The sum is not defined.
01
01,
654
321BA
Ch8.1-8.8_10
If k is a real number, then the scalar multiple is
DEFINITION 8.5Scalar Multiple of a Matrix
nmij
mnmm
n
n
ak
kakaka
kakaka
kakaka
k
)(
21
22221
11211
A
Ch8.1-8.8_11
If A, B, C are mn matrices, k1 and k2 are scalars(i) A + B = B + A(ii) A + (B + C) = (A + B) + C (iii) (k1k2) A = k1(k2A)
(iv) 1 A = A (v) k1(A + B) = k1A + k1B
(vi) (k1 + k2) A = k1A + k2A
THEOREM 8.1Matrix Multiplication
Ch8.1-8.8_12
If A is mp, B is pn, then the product is
DEFINITION 8.6Matrix Multiplication
pnpp
n
n
mpmm
p
p
bbb
bbb
bbb
aaa
aaa
aaa
21
22221
11211
21
22221
11211
AB
nm
p
kkjikba
1
Ch8.1-8.8_13
Example 3
(a)
(b)
Note: In general, AB BA
86
29,
53
74BA
3457
4878
65)2(36593
87)2(46794
........
AB
02
34,
72
01
85
BA
66
34
154
07)3(227)4(2
00)3(120)4(1
08)3(528)4(5
............
AB
Ch8.1-8.8_14
We can write a system of linear equations asthe form of product. eg:
Associate Law: A(BC) = (AB)C
Distributive Law: A(B + C) = AB + BC
(3) 83
24
83
24
21
21
2
1
xx
xx
x
x
Ch8.1-8.8_15
The transpose of (2) is
DEFINITION 8.7Transpose of a Matrix
mnnn
m
m
T
aaa
aaa
aaa
21
22212
12111
A
Ch8.1-8.8_16
Note: (A + B + C)T = AT + BT + CT (ABC)T = CTBTAT
Suppose A and B are matrices and k a scalar, (i) (AT)T = A(ii) (A + B)T = AT + BT (iii) (AB)T = BTAT
(iv) (kA)T = kAT
THEOREM 8.2Properties of Transpose
Ch8.1-8.8_17
Zero Matrices
A + 0 = A(4)
and A + (–A) = 0(5)
00
00
00
,00
00,
0
0000
Ch8.1-8.8_18
Triangular Matrices
For square n n matrices
TriangularLower TriangularUpper 143215
02111
00398
00061
00002
1000
9800
7650
4321
Ch8.1-8.8_19
Diagonal Matrices
For square n n matrices, i≠j, aij = 0
100
00
007
21
Ch8.1-8.8_20
Scalar Matrices
Diagonal matrices with equal aii
10
015
50
05
Ch8.1-8.8_21
Identity Matrices
A: m n, then Im A = A In = A
1000
0010
0001
Ch8.1-8.8_22
Symmetric
An n × n matrix A is said to be symmetric if AT = A.
467
652
721
A
AA
467
652
721T
Ch8.1-8.8_23
8.2 Systems of Linear Algebraic Equations
General Form
(1)mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111
Ch8.1-8.8_24
Solution
A solution of a (1) is said to be consistent if it has at least one solution and inconsistent if it has no solutions.
If a linear system is consistent:(i) a unique solution(ii) infinitely many solutions
See Fig 8.2
Ch8.1-8.8_25
Fig 8.2
Ch8.1-8.8_26
Example 1
Verify that x1 = 14 + 7t, x2 = 9 + 6t, x = t, where t is any real number, is a solution of
2x1 – 3x2 + x3 = 1 x1 – x2 – x3 = 5
Solution2(14 + 7t) – 3(9 + 6t) + 4t = 1 14 + 7t – (9 + 6t) – t = 5
For each number of t, we obtain a different solution.
2,3,0
4,33,42
0,9,14
321
321
321
xxx
xxx
xxx
txtxtx 321 ,69,714
Ch8.1-8.8_27
Solving Systems: Elementary Operations
(i) Multiply an equation by a nonzero constant.(ii) Interchange the position of equations(iii) Add a nonzero multiple of one equation to another one.
Ch8.1-8.8_28
Example 2
Solve 2x1 + 6x2 + x3 = 7 x1 + 2x2 – x3 = –1 5x1 + 7x2 – 4x3 = 9
Solution(i) Interchange positions
x1 + 2x2 – x3 = –1 … (a)2x1 + 6x2 + x3 = 7 … (b)
5x1 + 7x2 – 4x3 = 9 … (c)(ii) −2(a) + (b)
x1 + 2x2 – x3 = –1 … (a)2x2 + 3x3 = 9 … (b)’
5x1 + 7x2 – 4x3 = 9 … (c)
Ch8.1-8.8_29
Example 2 (2)
(iii) −5(a) + (c) x1 + 2x2 – x3 = –1 … (a)2x2 + 3x3 = 9 … (b)’
–3x2 + x3 = 14 … (c)’(iv) From (b)’ and (c)’, we have
x2 = −3, x3 = 5, then x1 = 10
Ch8.1-8.8_30
Augmented Matrix
To solve (1) we can use the augmented matrix
(2)
mmnmm
n
n
baaa
baaa
baaa
21
222221
111211
Ch8.1-8.8_31
Example 3
(a) The augmented matrix represents
x1 – 3x2 + 5x3 = 24x1 + 7x2 – x3 = 8
(b) x1 – 5x3 = – 1 x1 + 0x2 – 5x3 = – 12x1 + 8x2 = 7 and 2x1 + 8x2 + 0x3 = 7
x2 + 9x3 = 1 0x1 + x2 + 9x3 = 1are the same. Thus the matrix of the system is
8174
2531
1910
7082
1501
Ch8.1-8.8_32
Elementary Row Operations
(i) Multiply a row by a nonzero constant.(ii) Interchange the position of rows(iii) Add a nonzero multiple of one row to another one.
Matrices after elementary row operations are called row equivalent. The procedure is called row reduction.
Ch8.1-8.8_33
Example 4
(a) and
are in row-echelon form.
(b) and
are in reduced row-echelon form.
0000
1010
2051
410000
226100
0000
1010
7001
410000
606100
Ch8.1-8.8_34
Example 5
Solve the equations in example 2.
Solution(a)
9475
7162
1121
9475
1121
716212R
14130
10
1121
14130
9320
1121
29
23
52
221
3121
RRRRR
Ch8.1-8.8_35
Example 5 (2)
and
x3 = 5, x2 = –3, x1 = 10
5100
10
1121
00
10
1121
29
23
255
211
29
23
3 3112
32 RRR
529
23
12
3
32
321
x
xx
xxx
Ch8.1-8.8_36
Example 5 (3)
(b)
we have the same solution.
5100
3010
10001
5100
10
10401
5100
10
1121232
3134
29
23
122
29
23
RRRR
RR
Ch8.1-8.8_37
Example 6
Use Gauss-Jordan method to solvex1 + 3x2 – 2x3 = – 74x1 + x2 + 3x3 = 52x1 – 5x2 + 7x3 = 19
Solution
0000
3110
2101
3110
3110
7231
3311110
3311110
7231
19752
5314
7231
3212
3111
2111
3121
3
24
RRRR
RR
RRRR
Ch8.1-8.8_38
Example 6 (2)
We have x2 – x3 = –3 x1 + x3 = 2
Let x3 = t, then x2 = –3 + t, x1 = 2 – t.
Ch8.1-8.8_39
Example 7
Solve x1 + x2 = 14x1 − x2 = −62x1 – 3x2 = 8
Solution
We have 0 + 0 = 16, no solutions.
1600
210
101
832
614
111
Ch8.1-8.8_40
Networks
From Fig 8.3, we have
or(3) 0
0
0
3322
2211
321
RiRi
RiRiE
iii
0
0
3322
2211
321
RiRi
ERiRi
iii
Ch8.1-8.8_41
Fig 8.3
Ch8.1-8.8_42
Example 8
Solve (3) where R1 = 10 ohms, R2 = 20 ohms, R2 = 10 ohms, E = 12 volts
SolutionFrom the data, we have i1 – i2 – i3 = 0
10i1 + 20i2 = 1220i2 – 10i3 = 0
Ch8.1-8.8_43
Example 8 (2)
Use Gauss-Jordan method
We have i1 = 18/25, i2 = 6/25, i3 = 12/25.
2512
256
2518
operationsrow
100
010
001
010200
1202010
0111
Ch8.1-8.8_44
Homogeneous Systems
The system of equations
(4)
is always consistent, since x1 = x2 = … = xn = 0 will satisfy the system.
0
0
0
2211
2222112
1212111
nmnmm
nn
nn
xaxaxa
xaxaxa
xaxaxa
Ch8.1-8.8_45
The system (4) possesses nontrivial solutions if the
number m of equations is less than the number n of
unknowns.
THEOREM 8.3Existence of Nontrivial Solutions
Ch8.1-8.8_46
Example 9
Solve 2x1 − 4x2 + 3x3 = 0 x1 + x2 − 2x3 = 0
SolutionUse Gauss-Jordan method
0760
0211
0342
0211
0211
0342
21
12
2 RR
R
Ch8.1-8.8_47
Example 9 (2)
Let x3 = t, then x1 = (5/6)t, x2 = (7/6)t, and the system has nontrivial solutions.
010
001
010
0211
67
65
67
12
261
RR
R
Ch8.1-8.8_48
Example 10: Chemical Equations
Balance C2H6 + O2 CO2 + H2O
SolutionAssuming x1C2H6 + x2O2 x3CO2 + x4H2O, we have 2x1 = x3 (for C)
6x1 = 2x4 (for H)2x2 = 2x3 + 2x4 (for O)
Then
0100
0010
0001
01220
02006
00102
32
67
31
Ch8.1-8.8_49
Thus x4 = t, x1 = t/3, x2 = 7t/6, x3 = 2t/3. We choose t = 6, then
x1 = 2, x2 = 7, x3 = 4, x4 = 62C2H6 + 7O2 4CO2 + 6H2O
Example 10 (2)
Ch8.1-8.8_50
8.3 Rank of a Matrix
Introduction
Row vectors: u1 = (a11 a12 … a1n), u2 = (a21 a22, … a2n),…, um = (am1 am2 … amn)
mnmm
n
n
aaa
aaa
aaa
21
22221
11211
A
Ch8.1-8.8_51
Column Vectors:
mn
n
n
n
mm a
a
a
a
a
a
a
a
a
2
1
2
22
12
2
1
21
11
1 ,,, vvv
Ch8.1-8.8_52
The rank of a m n matrix A, denoted by rank(A), is
the maximum linearly independent row vectors.
DEFINITION 8.8Rank of a Matrix
Ch8.1-8.8_53
Example 1
Consider
(1)
The row vectors are in turn denoted as u1, u2, u3. Since 4u1 – ½u2 + u3 = 0, they are linearly dependent. In addition, u1 and u2 are not constant multiple couples, so they are linearly independent.
rank(A) = 2
8753
8622
3111
A
Ch8.1-8.8_54
Row Space
As in Example 1, Span(u1, u2, u3) is called the Row Space of A.
If a matrix A is row equivalent to a row-echelon form B, then(i) Row space of A = Row space of B(ii) The nonzero rows form a basis for the row
space of A(iii) rank(A) = the number of nonzero rows in B
THEOREM 8.4Rank of a Matrix by Row Reduction
Ch8.1-8.8_55
Example 2
Consider the matrix in (1)
rank(A) = 2
0000
210
3111
1420
2840
3111
8753
8622
3111
21
32
241
3221
31
21R
RRRRRR
A
Ch8.1-8.8_56
Example 3
Determine whether the set ofu1 = <2, 1, 1>, u2 = <0, 3, 0>, u3 = <3, 1, 2>
is linearly independent.
SolutionForm a matrix A by using u1, u2, u3, and reduce it.
We have rank(A) = 3, so the set of vectors is linearly independent.
100
010
001
213
030
112operations
row
A
Ch8.1-8.8_57
Rank and Linear Systems
Consider
Make an augmented matrix and reduce it:
832
64
1
21
21
21
xx
xx
xx
00
10
01
32
14
11operations
row
Ch8.1-8.8_58
We have rank(A|B) = 3. Since
then rank(A) = 2
100
010
001
1600
210
101
832
614
111
00
10
01
32
14
11operations
row
Ch8.1-8.8_59
AX = B is consistent if and only if rank(A|B) = rank(A)
THEOREM 8.5Consistence of AX = B
Suppose AX = B with m equations and n unknowns is
consistent. If rank(A) = r, then the solution contains
n – r parameters.
THEOREM 8.6Number of Parameters in a Solution
Ch8.1-8.8_60
Example
x1 + 3x2 – 2x3 = –7 4x1 + x2 + 3x3 = 5 (3)2x1 – 5x2 + 7x3 = 19
We already know that (3) is consistent and has infinitely many solutions. Since
We have rank(A|B) = rank(A) = 2, then the number of parameter of the solution is 3 – 2 = 1.
0000
3110
2101
19752
5314
7231operarions
row
Ch8.1-8.8_61
Flow chart
AX = 0
Always consistent
Unique solution X = 0
rank(A) = n
Infinity of solutions
Rank(A) < n
n – r parameters
Ch8.1-8.8_62
AX = B, B≠0
Inconsistent
rank(A) < rank(A│B)
consistent
rank(A) = rank(A│B)
Unique solution
rank(A) = nInfinity of solutions
rank(A) < n
n – r parameters
Ch8.1-8.8_63
8.4 Determinants
Notation
nnnn
n
n
aaa
aaa
aaa
21
22221
11211
A
nnnn
n
n
aaa
aaa
aaa
21
22221
11211
det A
Ch8.1-8.8_64
eg:
The determinant is
(1)
DEFINITION 8.9Determinant of 2 × 2 Matrix
2221
1211
aa
aaA
211222112221
1211det aaaaaa
aaA
695)3()9(695
36)det(,
95
36
AA
Ch8.1-8.8_65
The determinant is
(1)
DEFINITION 8.10Determinant of 3 × 3 Matrix
333231
232221
131211
aaa
aaa
aaa
A
.
det
332112322311
312213322113312312332211
333231
232221
131211
aaaaaa
aaaaaaaaaaaa
aaa
aaa
aaa
A
Ch8.1-8.8_66
In view of (1), we have
(4)
where A has been expanded by cofactors along the first row, with the cofactors of a11, a12, a13:
Thus det A = a11C11 + a12C12 + a13C13 (5)
3231
222113
3331
232112
3332
232211det
aa
aaa
aa
aaa
aa
aaa
A
3231
222113
3331
232112
3332
232211 aa
aaC
aa
aaC
aa
aaC
Ch8.1-8.8_67
In general, the cofactors of aij is Cij = (–1)i+ j Mij (6)
where Mij is called a minor determinant.From (3), we have
(8)
Similarly, we can expand by cofactors along the third rows:
det A = a31C31 + a32C32 + a33C33 (9)Note: We can expand by cofactors along any rows or any columns.
323222221212
2321
131132
3331
131122
3331
232112
211323113231133311223123332112 )()()(det
CaCaCa
aa
aaa
aa
aaa
aa
aaa
aaaaaaaaaaaaaaa
A
Ch8.1-8.8_68
Example 1
Find the determinant of
SolutionAlong the first row:
351
306
742
A
131211 742
351
306
742
det CCC A
35
30)1(
351
306
742
)1( 111111
C
Ch8.1-8.8_69
Example 1 (2)
31
36)1(
351
306
742
)1( 212112
C
51
06)1(
351
306
742
)1( 313113
C
120)]1(0)5(6[7)]1(3)3(6[4)]5(3)3(0[2
51
06)1(7
31
36)1(4
35
30)1(2det 312111
A
Ch8.1-8.8_70
We also can expanded along the second row, since it has zero entry.
120)6(3)23(6
51
42)1(3
35
74)1(6
306det
3221
232221
CCCA
Ch8.1-8.8_71
Example 2
Find the determinant of
SolutionAlong the third column:
238)]2(5)4(6[7
42
56)1)(7(
042
781
056
)1)(7(
0)7(0
042
781
056
det
3232
332313
CCCA
042
781
056
A
Ch8.1-8.8_72
Let A = (aij)n × n ne an n × n matrix. For each
the cofactor expansion of det A along the ith row is
For each the cofactor expansion of det A
along the ith column is
THEOREM 8.7Consistence of AX = B
,1 ni
,1 nj ininiiii CaCaCa 2211det A
njnjjjjj CaCaCa 2211det A
Ch8.1-8.8_73
Example 3
Find the determinant of
SolutionAlong the fourth row
where
4001
1611
3201
4215
A
44434241 )4(00)1(
4001
1611
3201
4215
det CCCC
A
Ch8.1-8.8_74
Example 3 (2)
161
320
421
)1( 1441
C
611
201
215
)1( 4444 C
1861
21)1(3
11
41)1(2
16
42)1(0
161
320
421
)1(
322212
41
C
Ch8.1-8.8_75
Example 3 (3)
411
15)1(2
61
25)1(0
61
21)1)(1(
611
201
215
322212
44
C
34)4)(4()18)(1(
4001
1611
3201
4215
det
A
Ch8.1-8.8_76
8.5 Properties of Determinants
If AT is the transpose of the n × n matrix A, then
det AT = det A
THEOREM 8.8Determinant of a Transpose
If any two rows (columns) of an n × n matrix A are the
same, then det A = 0.
THEOREM 8.9Two Identical Rows
4143
75det
A 41
47
35det
TA
Ch8.1-8.8_77
Example 1
229
224
226
A
0
229
224
226
det A
Ch8.1-8.8_78
If all the entries in a row (column) of an n × n matrix A
are all zeros, then det A = 0.
THEOREM 8.10Zero Row or Column
If B is the matrix obtained by interchanging any two
rows (columns) of an n × n matrix A, then
det B = −det A
THEOREM 8.11Interchanging Rows
Ch8.1-8.8_79
If B is obtained by interchanging the first and third row of
AB det
312
706
914
914
706
312
det
Ch8.1-8.8_80
If B is obtained from an n × n matrix A by multiplying
a row (column) by a nonzero real number k, then
det B = k det A
THEOREM 8.12Constant Multiple of a Row
A
B
A
det)(
det
rowth thealong cofactorsby det ofexpansion
2211
2211
kCaCaCak
CkaCkaCka
i
ininiiii
ininiiii
Ch8.1-8.8_81
Example 2
(a)
(b)
80)21(8012
11285
24
1185
164
815
1620
85
..
.
00)2(
227
115
114
)2(
247
125
124
.
Ch8.1-8.8_82
If A and B are both n × n matrices, then
det AB = det A det B.
THEOREM 8.13Determinant of a Matrix Product
Ch8.1-8.8_83
Example 3
Suppose
then
Now det AB = −24, det A = −8, det B = 3, Thus det AB = det A det B.
53
43,
11
62BA
96
2212AB
Ch8.1-8.8_84
Suppose B is the matrix obtained from an n × n matrix A by multiplying the entries in a row (column) by Nonzero real number k and adding the result to the corresponding entries in another row (column). Then det B = det A.
THEOREM 8.14Determinant is Unchanged
Ch8.1-8.8_85
Example 4
We have det A = 45 = det B = 45.
BA
2411
703
215
414
703
215313 RR
414
703
215
A
Ch8.1-8.8_86
Proof
Suppose A is an n × n matrix (upper or lower). Then
det A = a11 a22 … ann
where a11, a22, …, ann are the entries on the main
diagonal of A.
THEOREM 8.15Determinant of a Triangular Product
333231
2221
11
0
00
aaa
aa
a
A
332211323322113332
2211 )0(
0det aaaaaaa
aa
aa .A
Ch8.1-8.8_87
Example 5
(a)
2427
0495
0062
0003
A
144)2()4(63
2427
0495
0062
0003
det
...A
Ch8.1-8.8_88
(b)
400
060
003
A
7246)3(
400
060
003
det
..A
Ch8.1-8.8_89
Example 6
Find the determinant of
Solution
842
234
726
A
726
234
421
2
421
234
726
2
842
234
726
det
A
Ch8.1-8.8_90
Example 6 (2)
)19)(5)(1)(2(
1900
1850
421
2
17100
1850
421
2
726
1850
421
2
Ch8.1-8.8_91
Suppose A is a n n matrix. If ai1, ai2, …, ain are the
entries in the ith row and Ck1, Ck2, …, Ckn are the
cofactors in the kth row, then ai1 Ck1 + ai2 Ck2 + …+ ain Ckn = 0, for i k
If a1j, a2j, …, anj are the entries in the jth column and
C1k, C2k, …, Cnk are the cofactors in the kth column,
then a1j C1k + a2j C2k + …+ anj Cnk = 0, for j k
THEOREM 8.16A Property of Cofactors
Ch8.1-8.8_92
THEOREM 8.16
ProofLet B be the matrix obtained from A by letting the entries in the ith row of A be the same as the ones in the kth row, that is,
ai1 = ak1, ai2 = ak2, …, ain = akn then there are two identical rows in B, det B = 0,then
kninkiki
knknkkkk
CaCaCa
CaCaCa
2211
2211det0 B
Ch8.1-8.8_93
Example 7
Consider the matrix
we have
842
234
726
A
0)10(7)40(2)25(6
34
267
24
762
23
726
331332123111
CaCaCa
Ch8.1-8.8_94
8.6 Inverse of a Matrix
Note: If A has no inverse, then A is called singular.
Let A be an n n matrix. If there exists an n n matrixB such that
AB = BA = I (1)where I is the n n identity, then A is said to be nonsingular or invertible. Then B is said to be the inverse of A.
DEFINITION 8.11Inverse of a Matrix
Ch8.1-8.8_95
Proof(i) A-1A = AA-1 = I, A = (A-1)-1
(ii) (AB)(AB)-1 = I, B-1A-1(AB)(AB)-1 = B-1A-1
(AB)-1 = B-1A-1
Let A, B be nonsingular matrices. (i) (A-1)-1 = A (ii) (AB)-1 = B-1A-1 (iii) (AT)-1 = (A-1)T
THEOREM 8.17Properties of the Inverse
Ch8.1-8.8_96
Let A be an n × n matrix. The matrix that is the
Transpose of the matrix of cofactors corresponding to
the entries of A:
is called the adjoint of A and is denoted by adj A.
DEFINITION 8.12Adjoint Matrix
nnnn
n
nT
nnnn
n
n
CCC
CCC
CCC
CCC
CCC
CCC
21
22212
12111
21
22221
11211
Ch8.1-8.8_97
ProofFor n = 3,
(3)
Let A be an n × n matrix. If det A 0, then
(2)
THEOREM 8.18Finding the Inverse
AA
A adjdet
11
A
A
A
AA
det00
0det0
00det
)adj(
332313
322212
312111
333231
232221
131211
CCC
CCC
CCC
aaa
aaa
aaa
Ch8.1-8.8_98
From Theorem 8.16,
we have
Thus A-1 = adj A/det A
00
00
00
333322322131133312321131
332332223121132312221121
331332123111231322122111
CaCaCaCaCaCa
CaCaCaCaCaCa
CaCaCaCaCaCa
IAAAA )(det
100
010
001
)(det)adj(
Ch8.1-8.8_99
For 22 matrix
(4)
2221
1211
aa
aaA
1121
1222
1112
2122
2221
1211adjaa
aa
aa
aa
CC
CC TT
A
1121
12221
det1
aa
aa
AA
Ch8.1-8.8_100
For 33 matrix
(5)
3231
222113
3331
232112
3332
232211
333231
232221
131211
aa
aaC
aa
aaC
aa
aaC
aaa
aaa
aaa
A
332313
322212
3121111
det1
CCC
CCC
CCC
AA
Ch8.1-8.8_101
Example 1
Find the inverse of
Solution
Check
102
41A
21
1
1
25
12
410
21
A
10
01
541010
2245
1
25
102
41
21
1AA
10
01
5411
202045
102
41
1
25
21
1AA
Ch8.1-8.8_102
Example 2
Find the inverse of
SolutionSince
103
112
022
A
612
222
12
022
11
02
603
222
13
022
10
02
303
125
13
121
10
11
333231
232221
131211
CCC
CCC
CCC
Ch8.1-8.8_103
Example 2 (2)
We have
21
21
41
61
61
125
61
61
121
1
663
225
221
121
A
Ch8.1-8.8_104
An n n matrix A is nonsingular if and only if det A 0
THEOREM 8.19Nonsingular Matrices and det A
Ch8.1-8.8_105
Example 3
has no inverse.
A is singular, since det A = 0
33
22A
Ch8.1-8.8_106
An nn matrix A can be transformed into the n n identity matrix I by a sequence of elementary row operations, then A is nonsingular. The same operations that transforms A into the identity I will also transform I into A-1.
THEOREM 8.20Finding the Inverse
Ch8.1-8.8_107
First we construct the augmented matrix (A|I), and the process for finding A-1 is outlined.
100
010
001
)|(
21
22221
11211
nnnn
n
n
aaa
aaa
aaa
IA
Ch8.1-8.8_108
Row operations on A until I is obtained
)IA( | )A|I( 1
Simultaneously applying the same operations
Ch8.1-8.8_109
Example 4
Find the inverse of
Solution
655
432
102
A
1050
011530
0001
100655
010432
0001
100655
010432
001102
25
217
21
21
52
21
21
3121
121
RRRR
R
Ch8.1-8.8_110
Example 4 (2)
6105100
010
0001
00
010
0001
010
010
0001
31
31
35
21
21
30
51
31
61
301
31
31
35
21
21
51
21
1017
31
31
35
21
21
3
32
351
231
R
RR
RR
Ch8.1-8.8_111
Example 4 (3)
Thus
6105100
10178010
352001233
5132
1
RRRR
6105
10178
3521A
Ch8.1-8.8_112
Example 5
Find the inverse of
Solution
306
542
211
A
100306
012960
001211
100306
010542
001211
212 RR
Ch8.1-8.8_113
Example 5 (2)
There is a row of zeros, it is singular
114000
012960
001211
106960
012960
001211
32
316
RR
RR
Ch8.1-8.8_114
Using the Inverse
A system of m linear equations in n unknowns
(6)
can be written as AX = B, where
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111
,
21
22221
11211
mnmm
n
n
aaa
aaa
aaa
A ,2
1
nx
x
x
X
mb
b
b
2
1
B
Ch8.1-8.8_115
Special Case
When m = n, if A is nonsingular, then
X = A-1B(7)
Ch8.1-8.8_116
Example 6
Use the inverse to solve
SolutionThe system can be written as
Since , it is nonsingular. From (4)
1663
1592
21
21
xx
xx
16
15
63
92
2
1
x
x
03963
93
23
96
391
63
92 1
Ch8.1-8.8_117
Example 6 (2)
Using (7)
Thus
316
13
234
391
16
15
23
96
391
2
1
x
x
3/1,6 21 xx
Ch8.1-8.8_118
Example 7
Use the inverse to solve
SolutionFrom the equations we have
4432
1655
22
321
321
31
xxx
xxx
xx
655
432
102
A
Ch8.1-8.8_119
Example 7 (2)
Thus (7) gives
Thus
36
62
19
1
4
2
6105
10178
352
1
4
2
655
432
102 1
3
2
1
x
x
x
36,62,19 321 xxx
Ch8.1-8.8_120
A homogeneous system of n linear equations in n unknowns AX = 0 has only the trivial solution if and only if A is nonsingular.
THEOREM 8.21Trivial Solution Only
A homogeneous system of n linear equations in n unknowns AX = 0 has a nontrivial solution if and only if A is singular.
THEOREM 8.22Existence of Nontrivial Solutions
Ch8.1-8.8_121
8.7 Cramer’s Rule
IntroductionFor example, the equations
(1)
possesses the solution
(2)
where a11 a22 – a12 a21 0
2222121
1212111
bxaxa
bxaxa
,21122211
2122211 aaaa
baabx
21122211
2112112 aaaa
abbax
Ch8.1-8.8_122
Rewrite (2) as determinant forms, we have
(3)2221
1211
221
111
2
2221
1211
222
121
1 ,
aa
aaba
ba
x
aa
aaab
ab
x
Ch8.1-8.8_123
Special Matrix
For the following system
(4)nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111
Ch8.1-8.8_124
We define a special matrix
nnnknnknn
nkk
nkk
k
aabaaa
aabaaa
aabaaa
1121
2122122221
1111111211
A
Ch8.1-8.8_125
Let A be the coefficient matrix of the system (1).
If det A 0, then the solution of (1) is given by
where Ak, k = 1, 2, …, n, is defined in (5)
THEOREM 8.23Cramer’s Rule
,detdet
,,detdet
,detdet 2
21
1 AA
AA
AA n
nxxx
Ch8.1-8.8_126
Proof
nnnnn
nn
nn
nnnnn
n
n
cbCbCb
cbCbCb
cbCbCb
b
b
b
CCC
CCC
CCC
2211
2222121
1212111
2
1
21
22212
12111
det1
det1
A
ABAX 1-
Ch8.1-8.8_127
Now the entry in the kth row is
(7)
AA
A
detdet
det2211
k
nknkkk
CbCbCbx
Ch8.1-8.8_128
Example 1
Solve
Solution1245
33
723
321
321
321
xxx
xxx
xxx
,78
215
331
173
det
,13
245
311
123
det
2
A
A
52
145
311
723
det
39
241
313
127
det
3
1
A
A
Ch8.1-8.8_129
Example 1 (2)
From (3), we have
,3detdet 1
1 AA
x ,6detdet 2
2 AA
x 4detdet 3
3 AA
x
Ch8.1-8.8_130
8.8 The Eigenvalue Problems
Let A be n n matrix. A number is said to be an
eigenvalue of A if there exists a nonzero solution
vector K of AK = K (1)
The solution vector K is said to be an eigenvector
corresponding to the eigenvalue .
DEFINITION 8.13Eigenvalues and Eigenvectors
Ch8.1-8.8_131
Example 1
Verify that is an eigenvector of the matrix
SolutionSince
we conclude that K is an eigenvector of A.
1
1
1
K
112
332
310
A
KAK )2(
1
1
1
)2(
2
2
2
1
1
1
112
332
310eigenvalue
Ch8.1-8.8_132
From (1), we have(A – I)K = 0
(2)However, (2) is the same as a homogeneous system of linear equations. Since we want K to be nontrivial, we must have
det (A – I) = 0(4)Inspection of (4) shows det (A – I) results in an nth-degree polynomial in , and is called the characteristic equation.
Ch8.1-8.8_133
Example 2
Find the eigenvalues and eigenvectors of
Solution
121
016
121
A
0
121
016
121
)det(
IA
Ch8.1-8.8_134
Example 2 (2)
We have –3–2 + 12 = 0 or ( + 4)( – 3) = 0then = 0, −4, 3. To find the eigenvectors,(i) 1 = 0,
0000
010
001
0000
010
0121
0000
06130
0121
0121
0016
0121
)|0(
136
131
2
136
6
122131
3121
RRR
RRRR
0IA
Ch8.1-8.8_135
Example 2 (3)
Choose k3 = −13, then
3231 136
,131
kkkk
13
6
1
1K
Ch8.1-8.8_136
Example 2 (4)
(ii) For 2 = −4,
01680
01890
0321
0125
0036
0321
0321
0036
0125
)|4(
3121
133
56
RRRR
RR
0IA
Ch8.1-8.8_137
Example 2 (5)
implies k1 = −k3 , k2 = 2k3. Choose k3 = 1, then
0000
0210
0101
0210
0210
03213212
381
291
22
RRRR
RR
1
2
1
2K
Ch8.1-8.8_138
Example 2 (6)
(iii) 2 = 3,
implies k1 = – k3, k2 = –(3/2)k3. Choose k3 = –2,
0000
010
0101
0421
0046
0122
)|3( 23
operationsrow
0IA
2
3
2
3K
Ch8.1-8.8_139
Example 3
Find the eigenvalues and eigenvectors of
Solution
We see 1 = 2 = 5 is an eigenvalue of multiplicity 2.From (A – 5I|0), we get
71
43A
0)5(71
43)det( 2
IA
02
042
21
21
kk
kk
Ch8.1-8.8_140
Example 3 (2)
Choose k2 = 1, we have k1 = 2, then
We can have only one eigenvector though A is a 22 matrix.
1
21K
Ch8.1-8.8_141
Example 4
Find the eigenvalues and eigenvectors of
Solution
We see 1 = 11, 2 = 3 = 8 is of multiplicity 2.
911
191
119
A
0)8)(11(
911
191
119
)det( 2
IA
Ch8.1-8.8_142
Example 4 (2)
(i) For 1 = 11, Gauss-Jordan method gives
Hence k1 = k3, k2 = k3. If k3 = 1, then
0000
0110
0101
0211
0121
0112
)|11( 0IA
1
1
1
1K
Ch8.1-8.8_143
Example 4 (3)
(ii) Now for 2 = 8, we have
For k1 + k2 + k3 = 0, we can select two of them arbitrarily. We choose: k2 = 1, k3 = 0, and k2 = 0, k3 = 1, then
0000
0000
0111
0111
0111
0111
)|8( 30IA
,
0
1
1
2
K
1
0
1
3K
Ch8.1-8.8_144
ProofSince AK = K,The proof is completed.
Let A be a square matrix with real entries. If = + i,
0, is a complex eigenvalue of A, then its conjugate
is also an eigenvalue of A. If K is an
eigenvector corresponding to , then is an eigenvector
corresponding to .
THEOREM 8.24Complex Eignvalues and Eigenvectors
iK
,KKA KKA
Ch8.1-8.8_145
Example 5
Find the eigenvalues and eigenvectors of
Solution
For 1 = 5 + 2i,
45
16A
0291045
16)det( 2
IA
ii 25,25 121
0)21(5
0)21(
21
21
kik
kki
Ch8.1-8.8_146
Example 5 (2)
Since k2 = (1 – 2i) k1, after choosing k1 = 1, then
From Theorem 8.24, then
i21
11K
i21
112 KK
,2512 i
Ch8.1-8.8_147
The eigenvalues of an upper triangular, lower triangular,
or diagonal matrix are the main diagonal entries.
THEOREM 8.25Triangular and Diagonal Matrices