n unknowns x , x , , xn - UHjingqiu/math3321-2016F/3321-Ch5-slides.pdf · linear equation in two...

Chapter 5. Linear Algebra

A linear (algebraic) equation in

n unknowns, x1, x2, . . . , xn, is

an equation of the form

a1x1 + a2x2 + · · · + anxn = b

where a1, a2, . . . , an and b are

real numbers.

1

linear equation in one unknown

x: ax = b.

2

Exactly one of following holds:

(1) there is precisely one solution

x = b/a, a 6= 0,

(2) there are no solutions

0x = b, b 6= 0

(3) there are infinitely many solu-

tions

0x = 0.

3

linear equation in two unknowns

x, y:

ax + by = α.

A solution of the equation is an or-

dered pair of numbers (x, y).

If a = b = 0, and α = 0, then all

ordered pairs satisfy the equation.

If a = b = 0, and α 6= 0, then no

ordered pair satisfies the equation.

4

If a, b, not both 0, then the set

of all ordered pairs that satisfy the

equation is a straight line (in the

x, y-plane).

A system of two linear equations in

two unknowns:

ax + by = α

cx + dy = β

Find ordered pairs (x, y) that sat-

isfy both equations simultaneously.

5

Two lines in the plane either

(a) have a unique point of intersec-

tion (the lines have different slopes),

-4 -3 -2 -1 1 2 3

-2

2

4

6

8

10

12

6

(b) are parallel (the lines have the

same slope but, for example, differ-

ent y-intercepts)

-3 -2 -1 1 2 3

1

2

3

7

(c) coincide (same slope, same y-

intercept).

-3 -2 -1 1 2 3

1

2

8

That is, there is either a

(a) unique solution,

(b) no solution,

or

(c) infinitely many solutions.

9

A system of three linear equations

in two unknowns:

ax + by = α

cx + dy = β

ex + fy = γ

Find ordered pairs (x, y) that sat-

isfy the three equations simultane-

ously.

10

There is either a

(a) unique solution,

(b) no solution, (this is usually

what happens)

or

(c) infinitely many solutions.

11

Example:

x + y = 2−2x + y = 24x + y = 11

-1 1 2 3

5

10

12

A linear equation in three unknowns

x, y, z:

ax + by + cz = α.

A solution of the equation is an or-

dered triple of numbers (x, y, z).

If a = b = c = 0, and α = 0, all

ordered triples satisfy the equation.

If a = b = c = 0, and α 6= 0, no

ordered triple satisfies the equation.

13

If a, b, c, not all 0, then the set

of all ordered triples that satisfy the

equation is a plane (in 3-space).

-5 0 5

-5

0

5

-20

0

20

A system of two linear equations in

three unknowns

a11x + a12y + a13z = b1

a21x + a22y + a23z = b2

• Either the two planes are parallel

(the system has no solutions),

14

Figure

-2

0

2

-20

2

-40

-20

0

20

40

15

• they coincide (infinitely many so-

lutions, a whole plane of solutions),

• they intersect in a straight line

(again, infinitely many solutions.)

-5 0 5

-5

0

5

-20

0

20

16

A system of three linear equations

in three unknowns.

a11x + a12y + a13z = b1

a21x + a22y + a23z = b2

a31x + a32y + a33z = b3

The system represents three planes

in 3-space.

17

(a) The system has a unique solu-

tion; the three planes have a unique

point of intersection;

(b) The system has infinitely many

solutions; the three planes inter-

sect in a line, or the three planes

intersect in a plane.

(c) The system has no solution; there

is no point the lies on all three

planes.18

Systems of Linear Algebraic Equa-

tions

Example 1: Solve the system

x + 3y = −5

2x − y = 4

19

Equivalent system

x + 3y = −5

y = −2

Solution set:

x = 1, y = −2

20


x + 2y − 5z = −1−3x − 9y + 21z = 0

x + 6y − 11z = 1

21

Equivalent system

x + 2y − 5z = −1y − 2z = 1

z = −1

Solution set:

x = −4, y = −1, z = −1

22


3x − 4y − z = 32x − 3y + z = 1x − 2y + 3z = 2

23

Equivalent system

x − 2y + 3z = 2y − 5z = −3

0z = 1

The system has no solution.

24

The Elementary Operations

The operations that produce equiv-

alent systems are called elementary

operations.

1. Multiply an equation by a nonzero

number.

2. Interchange two equations.

3. Multiply an equation by a num-

ber and add it to another equation.

25


x1 − 2x2 + x3 − x4 = −2

−2x1 + 5x2 − x3 + 4x4 = 1

3x1 − 7x2 + 4x3 − 4x4 = −4

26

Equivalent system

x1 − 2x2 + x3 − x4 = −2

x2 + x3 + 2x4 = −3

x3 + 12x4 = −1

2

27

Solution set:

x1 = −132 − 3

2a,

x2 = −52 − 3

2a,

x3 = 12 − 1

2a,

x4 = a, a any real number.

28

Terms

A matrix is a rectangular array of

numbers. A matrix with m rows and

n columns is an m × n matrix.

Matrix representation of a sys-

tem of linear equations

a11x1 + a12x2 + · · · + a1nxn = b1a21x1 + a22x2 + · · · + a2nxn = b2

... ... ... ... ...

am1x1 + am2x2 + · · · + amnxn = bm

29

Augmented matrix and matrix of

coefficients:

Augmented matrix:

a11 a12 · · · a1n b1a21 a22 · · · a2n b2... ... ... ... ...

am1 am2 · · · amn bm

Matrix of coefficients:

a11 a12 · · · a1na21 a22 · · · a2n... ... ...

am1 a32 · · · amn

30

Elementary row operations:

1. Interchange row i and row j

Ri ↔ Rj.

2. Multiply row i by a nonzero

number k

kRi → Ri.

3. Multiply row i by a number k

and add the result to row j

kRi + Rj → Rj.

31

Examples

1. Solve the system

x + 2y − 5z = −1−3x − 9y + 21z = 0

x + 6y − 11z = 1

Augmented matrix:

1 2 −5 −1−3 −9 21 01 6 −11 1

32

Row reduce

1 2 −5 −1−3 −9 21 01 6 −11 1

33

1 2 −5 −10 1 −2 10 0 1 −1

Corresponding (equivalent) system

of equations:

x + 2y − 5z = −1

y − 2z = 1

z = −1

Solution set:

x = −4, y = −1, z = −1

34

2. Solve the system:

3x − 4y − z = 32x − 3y + z = 1x − 2y + 3z = 2

Augmented matrix:

3 −4 −1 32 −3 1 11 −2 3 2

.

35

Row reduce

3 −4 −1 32 −3 1 11 −2 3 2

36

Equivalent system

1 −2 3 20 1 −5 −30 0 0 1

Corresponding system of equations:

x − 2y + 3z = 20x + y − 5z = −3

0x + 0y + 0z = 1

That is

x − 2y + 3z = 2y − 5z = −3

0z = 1

Solution set: no solution.

37

3. Solve the system

x + y − 3z = 12x + y − 4z = 0

−3x + 2y − z = 7

Augmented matrix:

1 1 −3 12 1 −4 0

−3 2 −1 7

38

Row reduce

1 1 −3 12 1 −4 0

−3 2 −1 7

39

Equivalent system:

1 1 −3 10 1 −2 20 0 0 0

.


x + y − 3z = 10x + y − 2z = 2

0x + 0y + 0z = 0

or

x + y − 3z = 1y − 2z = 2

0z = 0

or

x + y − 3z = 1y − 2z = 2

40

This system has infinitely many so-

lutions given by:

x = −1 + a,

y = 2 + 2a,

z = a, a any real number.

41

Row echelon form:

1. Rows consisting entirely of ze-

ros are at the bottom of the matrix.

2. The first nonzero entry in a

nonzero row is a 1. This is called

the leading 1.

3. If row i and row i + 1 are

nonzero rows, then the leading 1 in

row i+1 is to the right of the leading

1 in row i.42

NOTE:

1. All the entries below a leading

1 are zero.

2. The number of leading 1’s is

less than or equal to the number of

rows.

3. The number of leading 1’s is


columns.

43

Solution method for systems of

linear equations:

1. Write the augmented matrix

(A|b) for the system.

2. Use elementary row operations

to transform the augmented matrix

to row echelon form.

3. Write the system of equa-

tions corresponding to the row ech-

elon form.44

4. Back substitute to find the

solution set.

This method is called Gaussian elim-

ination with back substitution.

45

Consistent/Inconsistent systems:

A system of linear equations is con-

sistent if it has at least one solu-

tion.

That is, a system is consistent if it

has either a unique solution or in-

finitely many solutions.

A system that has no solutions is

inconsistent.

46

Consistent systems:

A consistent system is said to be

independent if it has a unique so-

lution.

A system with infinitely many solu-

tions is called dependent.

47

4. Solve the system of equations

2x1 + 5x2 − 5x3 − 7x4 = 8x1 + 2x2 − 3x3 − 4x4 = 2

−3x1 − 6x2 + 11x3 + 16x4 = 0

Augmented matrix:

2 5 −5 −7 81 2 −3 −4 2

−3 −6 11 16 0

.

Transform to row echelon form:

48

Equivalent system:

1 2 −3 −4 20 1 1 1 40 0 1 2 3

.


x1 + 2x2 − 3x3 − 4x4 = 2

x2 + x3 + x4 = 4

x3 + 2x4 = 3

49

Solution set:

x1 = 9 − 4a,

x2 = 1 + a,

x3 = 3 − 2a,


50

5. Solve the system of equations

x1 − 3x2 + 2x3 − x4 + 2x5 = 2

3x1 − 9x2 + 7x3 − x4 + 3x5 = 7

2x1 − 6x2 + 7x3 + 4x4 − 5x5 = 7

Augmented matrix:

1 −3 2 −1 2 23 −9 7 −1 3 72 −6 7 4 −5 7

Transform to row echelon form:

51

Equivalent system:

1 −3 2 −1 2 20 0 1 2 −3 10 0 0 0 0 0

.


x1 − 3x2 + 2x3 − x4 + 2x5 = 20x1 + 0x2 + x3 + 2x4 − 3x5 = 1

0x1 + 0x2 + 0x3 + 0x4 + 0x5 = 0.

which is

x1 − 3x2 + 2x3 − x4 + 2x5 = 2x3 + 2x4 − 3x5 = 1

52

Solution set:

x1 = 3a + 5b − 8c,

x2 = a,

x3 = 1 − 2b + 3c,

x4 = b,

x5 = c,

a, b, c arbitrary real numbers

53

6. For what value(s) of k, if any,

does the system

x + y − z = 1

2x + 3y + kz = 3

x + ky + 3z = 2

have:

(a) a unique solution?

(b) infinitely many solutions?

(c) no solution?

54

Augmented matrix:

1 1 −1 12 3 k 31 k 3 2

55

1 1 −1 10 1 k + 2 10 0 (k + 3)(k − 2) k − 2

(a) Unique solution: k 6= 2,−3.

(b) Infinitely many solns: k = 2.

(c) No solution: k = −3.

56

If an m × n matrix A is reduced to

row echelon form, then the number

of non-zero rows in its row echelon

form is called the rank of A.

Equivalently, the rank of a matrix is

the number of leading 1’s in its row

echelon form.

57

Note:

1. The rank of a matrix is less

than or equal to the number of rows.

(Obvious)

2. The rank of a matrix is also


columns.

58

Consistent/Inconsistent Systems

Case 1: If the last nonzero row in

the row echelon form of the aug-

mented matrix is

(0 0 0 · · · 0 |1),

then that row corresponds to the

equation

0x1 + 0x2 + 0x3 + · · · + 0xn = 1,

which has no solutions. Therefore,

the system has no solutions.59

Note: In this case, rank of aug-

mented matrix > rank of coefficient

matrix

Case 2: If the last nonzero row has

the form

(0 0 0 · · · 1 ∗ · · · ∗ | b),

where the “1” is in the kth, k ≤ n

column, then the row corresponds

to the equation

0x1+· · ·+0xk−1+xk+(∗)xk+1+· · ·+(∗)xn = b

and the system either has a unique

solution or infinitely many solutions.

60

NOTE: In this case, rank of aug-

mented matrix = the rank of coef-

ficient matrix

61

Theorem: A system of linear equa-

tions is consistent if and only if the

rank of the coefficient matrix equals

the rank of the augmented matrix.

62

5.4. Reduced Row Echelon Form

Example Solve the system (c.f. Ex-

ample 1)

x + 2y − 5z = −1

−3x − 9y + 21z = 0

x + 6y − 11z = 1

Augmented matrix:

1 2 −5 −1−3 −9 21 01 6 −11 1

Row reduce to:

1 2 −5 −10 1 −2 10 0 1 −1

63

Corresponding (equivalent) system

of equations

x + 2y − 5z = −1

y − 2z = 1

z = −1

Back substitute to get:

x = −4, y = −1, z = −1.

64

Or, continue row operations:

1 2 −5 −10 1 −2 10 0 1 −1

→

1 0 0 −40 1 0 −10 0 1 −1

Corresponding system of equations

x = −4

y = −1

z = −1

65

Reduced Row Echelon Form

1. Rows consisting entirely of zeros

are at the bottom of the matrix.

2. The first nonzero entry in a

nonzero row is a 1.

3. The leading 1 in row i + 1 is to

the right of the leading 1 in row i.

4. The leading 1 is the only nonzero

entry in its column.

66

2. Solve the system (c.f. Example

4)

2x1 + 5x2 − 5x3 − 7x4 = 7

x1 + 2x2 − 3x3 − 4x4 = 2

−3x1 − 6x2 + 11x3 + 16x4 = 0

Augmented matrix:

2 5 −5 −7 71 2 −3 −4 2

−3 −6 11 16 0

.

Row echelon form:

1 2 −3 −4 20 1 1 1 30 0 1 2 3

.

67


x1 + 2x2 − 3x3 − 4x4 = 2

x2 + x3 + x4 = 3

x3 + 2x4 = 3

Solution set:

x1 = 11 − 4a,

x2 = a,

x3 = 3 − 2a,


68

Alternative solution: Reduced row

echelon form:

1 2 −3 −4 20 1 1 1 30 0 1 2 3

69

1 0 0 4 110 1 0 −1 00 0 1 2 3


x1 + 4x4 = 11

x2 − x4 = 0

x3 + 2x4 = 3

x1 = 11−4a, x2 = a, x3 = 3−2a, x4 = a,

a any real number.

70

The solution set can also be written

as:

x1x2x3x4

= a

−41

−21

+

11036

71

Homogeneous Systems

The system of linear equations

a11x1 + a12x2 + · · · + a1nxn = b1a21x1 + a22x2 + · · · + a2nxn = b2

... = ...am1x1 + am2x2 + · · · + amnxn = bm

is homogeneous if

b1 = b2 = · · · = bm = 0,

otherwise, the system is nonhomo-

geneous.

C.f. Linear differential equations.72

A homogeneous system

a11x1 + a12x2 + · · · + a1nxn = 0

a21x1 + a22x2 + · · · + a2nxn = 0... ... ... ... ...

am1x1 + am2x2 + · · · + amnxn = 0

ALWAYS has at least one solution,

namely

x1 = x2 = · · · = xn = 0,

called the trivial solution

That is, homogeneous systems are

always CONSISTENT.

73

3. Solve the homogeneous system

x − 2y + 2z = 0

4x − 7y + 3z = 0

2x − y + 2z = 0

Augmented matrix:

1 −2 2 04 −7 3 02 −1 2 0

Row echelon form:

1 −2 2 00 1 −5 00 0 1 0

74


x − 2y + 2z = 0

y − 5z = 0

z = 0.

This system has the unique solution

x = 0,

y = 0,

z = 0.

The trivial solution is the only solu-

tion.

75

4. Solve the homogeneous system

3x − 2y + z = 0

x + 4y + 2z = 0

7x + 4z = 0

Augmented matrix:

3 −2 1 01 4 2 07 0 4 0

Row echelon form:

1 4 2 00 1 5/14 00 0 0 0

76


x + 4y + 2z = 0

y +5

14z = 0

This system has infinitely many so-

lutions:

x = −4

7a,

y = −5

14a,

z = a, a any real number.

77

5.

2x1 + 5x2 − 5x3 + 7x4 + 5x5 = 2

x1 + 2x2 − 3x3 + 4x4 + 2x5 = 3

−3x1 − 6x2 + 9x3 − 10x4 − 2x5 = 0

Augmented matrix:

1 2 −3 4 2 32 5 −5 7 5 2

−3 −6 9 −10 −2 −3

.

78

Reduced row echelon form:

1 0 −5 0 −12 −70 1 1 −1 1 −40 0 0 1 2 3

79


x1 − 5x3 − 12x5 = −7

x2 + x3 + 3x5 = −1

x4 + 2x5 = 3

Solution set:

x1 = −7 + 5a + 12b,

x2 = −1 − a − 3b

x3 = a

x4 = 3 − 2b,

x5 = b a, b arbitrary real nos.

80

Write as an ordered quintuple:

x1x2x3x4x5

=

−7 + 5a + 12b−1 − a − 3b

a3 − 2b

b

= a

5−1100

+ b

12−30

−21

+

−7−1030

81

What can you say about

−7−1030

, and

5−1100

,

12−30

−21

?

82

a

5−1100

+ b

12−30

−21

is the set of all solutions of the ho-

mogeneous system

2x1 + 5x2 − 5x3 + 7x4 + 5x5 = 0

x1 + 2x2 − 3x3 + 4x4 + 2x5 = 0

−3x1 − 6x2 + 9x3 − 10x4 − 2x5 = 0

83

−7−1030

is a solution of the given

nonhomogeneous system

2x1 + 5x2 − 5x3 + 7x4 + 5x5 = 2

x1 + 2x2 − 3x3 + 4x4 + 2x5 = 3

−3x1 − 6x2 + 9x3 − 10x4 − 2x5 = 0

84

That is, the set of solutions of the

nonhomogeneous system consists of

• the set of all solutions of the cor-

responding homogeneous system

plus

• one solution of the nonhomoge-

neous system.

C.f. Section 3.4.

85

Section 5.5. Matrices and Vec-

tors

A matrix is a rectangular array of

objects arranged in rows and columns.

The objects are called the entries.

A matrix with m rows and n columns

is called an m × n matrix. m × n

is called the size of the matrix, and

the numbers m and n are its

dimensions.

86

A matrix in which the number of

rows equals the number of columns,

m = n, is called a square matrix

of order n.

87

If A is an m× n matrix, then aij

denotes the element in the ith row

and jth column of A:

A =

a11 a12 a13 · · · a1n

a21 a22 a23 · · · a2n

a31 a32 a33 · · · a3n... ... ... ...

am1 am2 am3 · · · amn

.

The notation A = (aij) also repre-

sents this display.

88

Special Cases: Vectors

A 1 × n matrix

v = (a1 a2 . . . an)

also written as

v = (a1, a2, . . . , an)

is called an row vector.

89

An m × 1 matrix

v =

a1a2...

am

is called a column vector.

The entries of a row or column vec-

tor are called the components of

the vector.

90

Arithmetic of Matrices

Let A = (aij) be an m×n matrix

and let B = (bij) be a p×q matrix.

• Equality: A = B if and only if

1. m = p and n = q;

2. aij = bij for all i and j.

That is, A = B if and only if A

and B are identical.

91

Example:

a b 32 c 0

=

7 −1 x2 4 0

if and only if

a = 7, b = −1, c = 4, x = 3.

92

• Addition: Let A = (aij) and

B = (bij) be m × n matrices.

A+B is the m×n matrix C = (cij)

where

cij = aij + bij for all i and j.

That is,

A + B = (aij + bij).

Addition of matrices is not de-

fined for matrices of different sizes.

93

Examples:

(a)

2 4 −32 5 0

+

−4 0 6−1 2 0

=

−2 4 31 7 0

(b)

2 4 −32 5 0

+

1 35 −30 6

is not defined.

94

Properties: Let A, B, and C be

matrices of the same size. Then:

1. A+B = B+A (Commutative)

2. (A + B) + C = A + (B + C)

(Associative)

A matrix with all entries equal to 0

is called a zero matrix. E.g.,

0 0 00 0 0

and

0 00 00 0

95

The symbol 0 will be used to de-

note the zero matrix of arbitrary size.

3. A + 0 = 0 + A = A.

The zero matrix is the additive iden-

tity.

96

The negative of a matrix A, de-

noted by −A is the matrix whose

entries are the negatives of the en-

tries of A.

4. A + (−A) = 0.

Subtraction: Let A = (aij) and

B = (bij) be m × n matrices. Then

A − B = A + (−B).

97

Example:

2 4 −32 5 0

−

−4 0 6−1 2 0

=

6 4 −93 3 0

.

98

• Multiplication of a Matrix by a

Number

The product of a number k and

a matrix A, denoted kA, is given

by

kA = (kaij).

This product is also called multipli-

cation by a scalar.

99

Example:

−3

2 −1 41 5 −24 0 3

=

−6 3 −12−3 −15 6−12 0 −9

.

100

Properties: Let A, B be m × n

matrices and let α, β be real num-

bers. Then

1. 1A = A

2. 0A = 0

3. α(A + B) = α A + α B

4. (α + β)A = α A + β A

101

Matrix Multiplication

1. The Product of a Row Vector

and a Column Vector: The prod-

uct of a 1 × n row vector and an

n×1 column vector is the number

given by

(a1, a2, a3, . . . , an)

b1b2b3...bn

= a1b1 + a2b2 + a3b3 + · · · + anbn.

102

Also called scalar product (because

the result is a number (scalar)), dot

product, and inner product.

The product of a row vector and a

column vector (of the same dimen-

sion and in that order!) is a num-

ber.

The product of a row vector and

a column vector of different dimen-

sions is not defined.

103

Examples

(3, −2, 5)

−1−41

= 3(−1) + (−2)(−4) + 5(1) = 10

(−2, 3, −1, 4)

24−3−5

= −2(2)+3(4)+(−1)(−3)+4(−5) = −9

104

2. Matrix Multiplication: Let

A = (aij) be an m × p matrix and

let B = (bij) be a p × n matrix.

The matrix product of A and B

(in that order), denoted AB, is the

m×n matrix C = (cij), where cij

is the product of the ith row of A

and the jth column of B.

105

Let A and B be given matrices.

The product AB, in that order, is

defined if and only if the number of

columns of A equals the number

of rows of B.

If the product AB is defined, then

the size of the product is: (no. of

rows of A)×(no. of columns of B):

Am×p

Bp×n

= Cm×n

106

Examples:

1. A =

1 4 23 1 5

, B =

3 0−1 21 −2

AB:

AB =

1 4 23 1 5

3 0−1 21 −2

=

1 413 −8

BA:

3 0−1 21 −2

1 4 23 1 5

=

3 12 65 −2 8−5 2 −8

.

107

2. A =

1 23 4

, B =

1 −1 2−3 0 5

AB =

−5 −1 12−9 −3 26

BA does not exist.

3. A =

2 1−4 −2

, B =

1 1−2 −2

AB =

0 00 0

, BA =

−2 −14 2

Conclusion: Matrix multiplication

is not commutative; AB 6= BA in

general.

108

Properties of Matrix Multiplica-

tion: Let A, B, and C be ma-

trices.

1. AB 6= BA in general; NOT

COMMUTATIVE.

2. (AB)C = A(BC) matrix multi-

plication is associative.

109

Identity Matrices: Let A be a

square matrix of order n. The en-

tries a11, a22, a33, . . . , ann form

the main diagonal of A.

For each positive integer n > 1,

let In denote the square matrix of

order n whose entries on the main

diagonal are all 1, and all other

entries are 0. The matrix In is

called the n× n identity matrix.

110

I2 =

1 00 1

,

I3 =

1 0 00 1 00 0 1

,

I4 =

1 0 0 00 1 0 00 0 1 00 0 0 1

and so on.

111

3. If A is an m×n matrix, then

ImA = A and AIn = A.

If A is a square matrix of order n,

then

AIn = InA = A,

4. Inverse ?????

112

Distributive Laws:

1. A(B + C) = AB + AC. This is

called the left distributive law.

2. (A+ B)C = AC + BC. This is

called the right distributive law.

3. k(AB) = (kA)B = A(kB)

113

Quiz 4

A =

2 3 −1−4 2 05 −8 2

, B =

3 −14 27 6

If C = AB, then

(a) c22 =

(b) c13 =

114

Other ways to look at systems of

linear equations.

A system of m linear equations in n

unknowns:

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

a31x1 + a32x3 + · · · + a3nxn = b3

.............................................

am1x1 + am2x2 + · · · + amnxn = bm

115

Because of the way we defined mul-

tiplication,

a11 a12 a13 · · · a1n

a21 a22 a23 · · · a2n

a31 a32 a33 · · · a3n... ... ... ... ...


x1

x2

x3...

xn

=

b1b2b3...

bm

or in the vector-matrix form

Ax = b c.f. ax = b. (1)

Solution: x = A−1b ?????

116

Square matrices

1. Inverse

Let A be an n × n matrix. An

n × n matrix B with the property

that

AB = BA = In

is called the multiplicative inverse

of A or, more simply, the inverse

of A.

117

Uniqueness: If A has an inverse,

then it is unique. That is, there is

one and only one matrix B such

that

AB = BA = I.

B is denoted by A−1.

118

Procedure for finding the inverse:

A =

1 23 4

We want

1 23 4

x yz w

=

1 00 1

119

Examples:

1. A =

1 23 4

1 2 1 03 4 0 1

−→

1 0 −2 1

0 1 32 −1

2

A−1 =

−2 132 −1

2

120

2. B =

2 −1−4 2

2 −1 1 0−4 2 0 1

B does not have an inverse.121

NOTE: Not every nonzero n × n

matrix A has an inverse!

122

3. C =

1 0 22 −1 34 1 8

Form the augmented matrix:

1 0 2 1 0 02 −1 3 0 1 04 1 8 0 0 1

123

1 0 2 1 0 02 −1 3 0 1 04 1 8 0 0 1

−→

1 0 0 −11 2 2

0 1 0 −4 0 1

0 0 1 6 −1 −1

C−1 =

−11 2 2−4 0 16 −1 −1

124

Finding the inverse of A. Let A

be an n × n matrix.

a. Form the augmented matrix (A|In).

b. Reduce (A|In) to reduced row

echelon form. If the reduced row

echelon form is

(In|B), then B = A−1

If the reduced row echelon form is

not (In|B), then A does not have

an inverse.125

Application: Solve the system

x +2z = 32x −y +3z = 54x +y +8z = −2

Written in matrix form:

1 0 22 −1 34 1 8

xyz

=

35

−2

Solution: Cx = b; x = C−1b

xyz

=

−11 2 2−4 0 16 −1 −1

35

−2

=

−29−1415

126

2. Determinants

A. Calculation

1. 2 × 2

∣

∣

∣

∣

∣

∣

a bc d

∣

∣

∣

∣

∣

∣

= ad − bc.

127

2. 3 × 3

∣

∣

∣

∣

∣

∣

∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣

∣

∣

∣

∣

∣

∣

=

a1b2c3−a1b3c2−a2b1c3+a2b3c1+a3b1c2−a3b2c1

a1(b2c3−b3c2)−a2(b1c3−b3c1)+a3(b1c2−b2c1)

128

= a1

∣

∣

∣

∣

∣

∣

b2 b3c2 c3

∣

∣

∣

∣

∣

∣

−a2

∣

∣

∣

∣

∣

∣

b1 b3c1 c3

∣

∣

∣

∣

∣

∣

+a3

∣

∣

∣

∣

∣

∣

b1 b2c1 c2

∣

∣

∣

∣

∣

∣

This is called the expansion across

the frist row.

129

∣

∣

∣

∣

∣

∣

∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣

∣

∣

∣

∣

∣

∣

=


= −b1(a2c3−a3c2)+b2(a1c3−a3c1)−b3(a1c2−a2c1)

= −b1

∣

∣

∣

∣

∣

a2 a3c2 c3

∣

∣

∣

∣

∣

+ b2

∣

∣

∣

∣

∣

a1 a3c1 c3

∣

∣

∣

∣

∣

− b3

∣

∣

∣

∣

∣

a1 a2c1 c2

∣

∣

∣

∣

∣

Expansion across the second row.

130

∣

∣

∣

∣

∣

∣

∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣

∣

∣

∣

∣

∣

∣

=


= a3b1c2−a3b2c1−b3a1c2+b3a2c1+c3a1b2−c3a2b1

= a3(b1c2−b2c1)−b3(a1c2−a2c1)+c3(a1b2−a2b1)

= a3

∣

∣

∣

∣

∣

b1 b2c1 c2

∣

∣

∣

∣

∣

− b3

∣

∣

∣

∣

∣

a1 a2c1 c2

∣

∣

∣

∣

∣

+ c3

∣

∣

∣

∣

∣

a1 a2b1 b2

∣

∣

∣

∣

∣

Expansion down the third column

131

and so on. You can expand across

any row, or down any column.

BUT: Associated with each position

is an algebraic sign:∣

∣

∣

∣

∣

∣

∣

∣

+ − +− + −+ − +

∣

∣

∣

∣

∣

∣

∣

∣

For example, across the second row:

∣

∣

∣

∣

∣

∣

∣

∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣

∣

∣

∣

∣

∣

∣

∣

=

= −b1

∣

∣

∣

∣

∣

∣

a2 a3c2 c3

∣

∣

∣

∣

∣

∣

+b2

∣

∣

∣

∣

∣

∣

a1 a3c1 c3

∣

∣

∣

∣

∣

∣

−b3

∣

∣

∣

∣

∣

∣

a1 a2c1 c2

∣

∣

∣

∣

∣

∣

132

3. 4 × 4 determinants∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

a1 a2 a3 a4b1 b2 b3 b4c1 c2 c3 c4d1 d2 d3 d4

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

Sign chart∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

+ − + −− + − ++ − + −− + − +

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

133

Cramer’s Rule.

Given a system of n linear equations

in n unknowns: (a “square” system)

a11x1 + a12x2 + · · · + a1nxn = b1a21x1 + a22x2 + · · · + a2nxn = b2a31x1 + a32x2 + · · · + a3nxn = b3.............................................

an1x1 + an2x2 + · · · + annxn = bn

xi =detAi

detA, (provided detA 6= 0)

where Ai is the matrix A with the

ith column replaced by the vector

b.134

If

detA 6= 0,

then the system has a unique solu-

tion.

If

detA = 0,

then the system either has infinitely

many soluitons or no solutions.

135

Examples:

1. Given the system

x + 2y + 2z = 3

2x − y + z = 5

−4x + y − 2z = −2

Does Cramer’s rule apply?

det A =

∣

∣

∣

∣

∣

∣

∣

∣

1 2 22 −1 1−4 1 −2

∣

∣

∣

∣

∣

∣

∣

∣

= −3

Yes! Cramer’s Rule applies:

136

x + 2y + 2z = 3

2x − y + z = 5

−4x + y − 2z = −2

Find y.

y =

∣

∣

∣

∣

∣

∣

∣

∣

1 3 22 5 1−4 −2 −2

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

1 2 22 −1 1−4 1 −2

∣

∣

∣

∣

∣

∣

∣

∣

=−22

3

137

2. Given the system

−2x + 7y + 6z = −1

5x + y − 2z = 7

3x + 8y + 4z = −1

det A =

∣

∣

∣

∣

∣

∣

∣

∣

−2 7 65 1 −23 8 4

∣

∣

∣

∣

∣

∣

∣

∣

= 0

Cramer’s Rule does not apply

Does the system have infinitely many

solutions or no solutions??

Compare with ax = b.

138

B. Properties of determinants:

Let A be an n × n matrix.

1. If A has a row or column of

zeros, then det A = 0

Example:

1 0 22 0 34 0 8

Expand down second column:

−0

∣

∣

∣

∣

∣

∣

2 34 8

∣

∣

∣

∣

∣

∣

+ 0

∣

∣

∣

∣

∣

∣

1 24 8

∣

∣

∣

∣

∣

∣

− 0

∣

∣

∣

∣

∣

∣

1 23 2

∣

∣

∣

∣

∣

∣

= 0

139

2. If A is a diagonal matrix,

A =

a1 0 00 b2 00 0 c3

,

det A = a1

∣

∣

∣

∣

∣

∣

b2 00 c3

∣

∣

∣

∣

∣

∣

−0

∣

∣

∣

∣

∣

∣

0 00 c3

∣

∣

∣

∣

∣

∣

+0

∣

∣

∣

∣

∣

∣

0 b20 0

∣

∣

∣

∣

∣

∣

= a1 · b2 · c3.

In particular, det In = 1

For example, I3:∣

∣

∣

∣

∣

∣

∣

∣

1 0 00 1 00 0 1

∣

∣

∣

∣

∣

∣

∣

∣

= 1

140

3. If A is a triangular matrix,

e.g.,

A =

a1 a2 a3 a40 b2 b3 b40 0 c3 c40 0 0 d4

(upper triangular)

Expand down first column!!

Then det A = a1 · b2 · c3 · d4

141

4. If B is obtained from by inter-

changing any two rows (columns),

then

detB = −detA.

142

NOTE: If A has two identical rows

(or columns), then

detA = 0.

143

5. Multiply a row (column) of A

by a nonzero number k to obtain

a matrix B. Then

detB = k detA.

144

6. Multiply a row (column) of

A by a number k and add it to

another row (column) to obtain a

matrix B. Then

detB = detA.

145

7. Let A and B be n × n

matrices. Then

det [AB] = detAdetB.

146

Example:

Calculate

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

3 3 1 52 2 0 −24 1 −3 22 10 3 2

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

147

= −6

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

1 1 0 −10 1 1 −20 0 1 80 0 0 60

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

= −360

148

Equivalences:

1. The system of equations:

Ax = b has a unique solution.

2. The reduced row echelon form

of A is In.

3. The rank of A is n.

4. A has an inverse.

5. det A 6= 0.

149

A is nonsingular if det A 6= 0;

A is singular if det A = 0.

150

Linear Dependence/Independence

in Rn

R2 = {(a, b) : a, b ∈ R} – ”the plane”

R3 = {(a, b, c) : a, b, c ∈ R} – ”3-

space”

R4 = {(x1, x2, x3, x4)} – “4-space”

and so on

151

Let S = {v1,v2, · · · ,vk} be a set

of vectors in Rn.

The set S is linearly dependent if

there exist k numbers c1, c2, · · · , ck

NOT ALL ZERO such that

c1v1 + c2v2 + · · · + ckvk = 0.

(c1v1 + c2v2 + · · ·+ ckvk is a linear

combination of v1, v2, . . .)

152

S is linearly independent if it is

not linearly dependent. That is, S

is linearly independent if

c1v1 + c2v2 + · · · + ckvk = 0

implies c1 = c2 = · · · = ck = 0.

153


there exist k numbers c1, c2, · · · , ck


c1v1 + c2v2 + · · · + ckvk = 0.

Another way to say this:


one of the vectors can be written as

a linear combination of the others.

154

1. Two vectors v1, v2.

Linearly dependent iff one vector is

a multiple of the other.

Examples:

v1 = (1, −2, 4 ), v2 = (−1

2, 1, −2 )

linearly dependent: v1 = −2v2

155

v1 = (2, −4, 5 ), v2 = (0, 0, 0 )

linearly dependent: v2 = 0v1

v1 = (5, −2, 0 ), v2 = (−3, 1, 9 )

linearly independent

156

2. v1 = (1, −1 ), v2 = (2, −3 )

v3 = (3, −5 )

{v1, v2}

Dependent or independent??

{v1, v2, v3}


157

Does there exist three numbers, c1, c2, c3,

not all zero such that

c1v1 + c2v2 + c3v3 = (0,0)

That is, does the system of equa-

tions

c1 + 2c2 + 3c3 = 0

−c1 − 3c2 − 5c3 = 0

have nontrivial solutions?

158

A homogeneous system with more

unknowns than equations always

has infinitely many nontrivial so-

lutions.

Let v1,v2, · · · ,vk be a set of k

vectors in Rn. If k > n, then the

set of vectors is (automatically)

linearly dependent.

159

3.

v1 = (1, −1, 2 ), v2 = (2, −3, 0 ),

v3 = (−1, −2, 2 ), v4 = (0, 4, −3 )


a. {v1, v2, v3, v4}

b. {v1, v2, v3}

c. {v1, v2}

160

v1 = (1, −1, 2 ), v2 = (2, −3, 0 ),

v3 = (−1, −2, 2 )

Does c1v1 + c2v2 + c3v3 = 0

have non-trivial solutions? That is,

does

c1 + 2c2 − c3 = 0

−c1 − 3c2 − 2c3 = 0

2c1 + 2c3 = 0

have non-trivial solutions

161

Augmented matrix and row reduce:

1 2 −1 0−1 −3 −2 02 0 2 0

162

NOTE: It’s enough to row reduce:

1 2 −1−1 −3 −22 0 2

163

Calculate the determinant∣

∣

∣

∣

∣

∣

∣

∣

1 2 −1−1 −3 −22 0 2

∣

∣

∣

∣

∣

∣

∣

∣

164

Or, row reduce

1 −1 22 −3 0

−1 −2 2

165

4. v1 = ( a, 1, −1 ), v2 = (−1, 2a, 3 ),

v3 = (−2, a, 2 ), v4 = (3a, −2, a )

For what values of a are the vectors

linearly dependent?

166

4. v1 = ( a, 1, −1 ), v2 = (−1, 2a, 3 ),

v3 = (−2, a, 2 ), v4 = (3a, −2, a )

For what values of a are the vectors

v1, v2, v3 linearly dependent?

a c1 − c2 − 2c3 = 0

c1 + 2a c2 + a c3 = 0

−c1 + 3c2 + 2c3 = 0

167

Augmented matrix and row reduce:

a −1 −2 01 2a a 0

−1 3 2 0

168

Or row reduce:

a −1 −21 2a a

−1 3 2

169

Calculate the determinant∣

∣

∣

∣

∣

∣

∣

∣

a 1 −1−1 2a 3−2 a 2

∣

∣

∣

∣

∣

∣

∣

∣

170

5. v1 = (1,−1,2,1), v2 = (3,2,0,−1)

v3 = (−1,−4,4,3), v4 = (2,3,−2,−2)

a. {v1, v2, v3, v4}

dependent or independent?

b. What is the maximum number

of independent vectors?

171

Row reduce

1 −1 2 13 2 0 −1

−1 −4 4 32 3 −2 −2

172

Or, calculate the determinant.∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

1 −1 2 13 2 0 −1

−1 −4 4 32 3 −2 −2

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

173

Tests for independence/dependence

Let S = v1,v2, · · · ,vk be a set of

vectors in Rn.

Case 1: k > n : S is linearly

dependent.

174

Case 2: k = n : Form the n × n

matrix A whose rows are v1, v2, · · · ,vn

1. Row reduce A:

if the reduced matrix has n nonzero

rows –independent;

one or more zero rows – depen-

dent.

175

Equivalently, solve the system of

equations

c1v1 + c2v2 + · · · + ckvk = 0.

Unique solution: c1 = c2 = · · · =

cn = 0 – independent; infinitely

many solutions – dependent.

2. Calculate det A:

det A 6= 0 –independent;

det A = 0 – dependent.

Note: If v1,v2, · · · ,vn is a lin-

early independent set of vectors in

Rn, then each vector in Rn has

a unique representation as a linear

combination of v1,v2, · · · ,vn.

Case 3: k < n : Form the k×n ma-

trix A whose rows are v1, v2, · · · ,vk

1. Row reduce A:

if the reduced matrix has k nonzero

rows –independent;

one or more zero rows – depen-

dent.

176

Equivalently, solve the system of

equations

c1v1 + c2v2 + · · · + ckvk = 0.

Unique solution: c1 = c2 = · · · =

cn = 0 – independent; infinitely

many solutions – dependent.

Linear Dependence/Independence

of Functions

Let S = {f1(x), f2(x) · · · , fk(x)}

be a set of functions defined on an

interval I.

S is linearly dependent if there

exist k numbers c1, c2, · · · , ck


c1f1(x)+c2f2(x)+· · ·+ckfk(x) ≡ 0 on I.

177

S is linearly independent if it is

not linearly dependent. That is, S

is linearly independent if

c1f1(x)+c2f2(x)+· · ·+ckfk(x) ≡ 0 on I

implies c1 = c2 = · · · = ck = 0.

Examples:

178

1.

f1(x) = 1, f2(x) = x, f3(x) = x2

Suppose f1, f2, f3 are linearly de-

pendent. Then there exist 3 num-

bers c1, c2, c3, NOT ALL 0 such

that

c1 · 1 + c2x + c3x2 ≡ 0

That is,

c1 · 1 + c2x + c3x2 = 0

for all x179

2. f1(x) = sin x, f2(x) = cos x

Suppose f1, f2 are linearly depen-

dent. Then there exist 2 numbers

c1, c2, NOT BOTH 0 such that

c1 cos x + c2 sin x ≡ 0

That is,

c1 cos x + c2 sin x = 0

for all x

180

3. f1(x) = sin x, f2(x) = cos x,

f3(x) = cos (x + π/3)

181

Test for independence/dependence

Suppose that the functions f1(x), f2(x) · · · , fk(x)

are (k−1)-times differentiable on I. If the

determinant

W (x) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

f1(x) f2(x) · · · fk(x)

f ′1(x) f ′

2(x) · · · f ′k(x)

· · · · · ·

f(k−1)1 (x) f

(k−1)2 (x) · · · f

(k−1)k (x)

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

6= 0

for at least one x ∈ I, then the functions

are linearly independent on I.

NOTE: If W (x) ≡ 0 in I, then ??????

The functions may be dependent, they

may be independent!!!!

182

W (x) is called THEEEEEE

WRONSKIAN

183

Examples:

f1 = 2x+1, f2 = x−3, f3 = 3x+2

Wronskian:∣

∣

∣

∣

∣

∣

∣

∣

2x + 1 x − 3 3x + 22 1 30 0 0

∣

∣

∣

∣

∣

∣

∣

∣

= 0

Independent or dependent???

Does there exist 3 numbers, c1, c2, c3,

not all 0, such that

c1f1 + c2f2 + c3f3 ≡ 0

184

f1(x) = x3, f2(x) =

−x3, x < 0

2x3, x ≥ 0

Wronskian:

For x < 0

∣

∣

∣

∣

∣

∣

x3 −x3

3x2 −3x2

∣

∣

∣

∣

∣

∣

= 0

For x ≥ 0

∣

∣

∣

∣

∣

∣

x3 2x3

3x2 6x2

∣

∣

∣

∣

∣

∣

= 0

185

Suppose there exist 2 numbers c1, c2,

not both zero, such that

c1f1(x) + c2f2(x) ≡ 0

186

Interpretations of a system of linear equa-

tions

Given the system of m linear equations in

n unknowns:

a11x1 + a12x2 + a13x3 + · · · + a1nxn = b1

a21x1 + a22x2 + a23x3 + · · · + a2nxn = b2

a31x1 + a32x2 + a33x3 + · · · + a3nxn = b3

.........................................................

am1x1 + am2x2 + am3x3 + · · · + amnxn = bm

187

The coefficient matrix A is a mapping

from Rn to R

m

a11 a12 a13 · · · a1n

a21 a22 a23 · · · a2n

a31 a32 a33 · · · a3n... ... ... ...


x1

x2

x3...

xn

=

b1b2b3...

bm

188

A is a linear transformation!

A[x + y] = Ax + Ay and A[αx] = α Ax

Example:

A =

(

1 2 −13 0 2

)

is a linear transformation from R3 to R2.

189

Write the system as

a11 a12 a13 · · · a1n

a21 a22 a23 · · · a2n

a31 a32 a33 · · · a3n... ... ... ...


x1

x2

x3...

xn

=

b1b2b3...

bm

or in the vector-matrix form

Ax = b

Given a vector b in Rm, find a vector x

in Rn such that Ax = b.

190

Write the system equivalently as

x1

a11a21

...am1

+x2

a12a22

...am2

+· · ·+xn

a1na2n

...amn

=

b1b2...

bm

This says, express b as a linear combina-

tion of the columns of A.

191

Eigenvalues/Eigenvectors

Example: Set

A =

1 −3 1−1 1 13 −3 −1

.

A is a linear transformation from

R3 to R3.

1 −3 1−1 1 13 −3 −1

111

=

−11

−1

192

1 −3 1−1 1 13 −3 −1

1−32

=

12−213

1 −3 1−1 1 13 −3 −1

32

−3

=

−6−46

= −2

32

−3

1 −3 1−1 1 13 −3 −1

101

=

202

= 2

101

193

Let A be an n × n matrix.

A number λ is an eigenvalue of

A if there is a non-zero vector v

such that

Av = λv

194

To find the eigenvalues of A, find

the values of λ that satisfy

det (A − λ I) = 0.

Example: Let A =

2 31 4

A − λI =

2 − λ 31 4 − λ

det (A − λI) =

∣

∣

∣

∣

∣

∣

2 − λ 31 4 − λ

∣

∣

∣

∣

∣

∣

= (2−λ)(4−λ)−3 = λ2−6λ+5 = 0

Eigenvalues: λ1 = 5, λ2 = 1

195

Let A =

1 −3 1−1 1 13 −3 −1

det (A−λI) =

∣

∣

∣

∣

∣

∣

∣

∣

1 − λ −3 1−1 1 − λ 13 −3 −1 − λ

∣

∣

∣

∣

∣

∣

∣

∣

= −λ3 + λ2 + 4λ − 4

det (A − λI) = 0 implies

−λ3 + λ2 + 4λ − 4 = 0

or λ3 − λ2 − 4λ + 4 = 0

196

Terminology:

• det (A − λ I) is a polynomial of

degree n, called the characteristic

polynomial of A.

• The zeros of the characteristic poly-

nomial are the eigenvalues of A

• The equation det (A−λ I) = 0 is

called the characteristic equation

of A.

197

• A non-zero vector v that satisfies

Av = λv

is called an eigenvector correspond-

ing to the eigenvalue λ.

Examples: 1. A =

2 22 −1

Characteristic polynomial:

det (A − λ I) =

∣

∣

∣

∣

∣

∣

2 − λ 22 −1 − λ

∣

∣

∣

∣

∣

∣

= (2 − λ)(−1 − λ) − 4

= λ2 − λ − 6

Characteristic equation:

λ2 − λ − 6 = (λ − 3)(λ + 2) = 0

Eigenvalues: λ1 = 3, λ2 = −2

198

Eigenvectors:

199

2. A =

4 −14 0


det (A − λ I) =

∣

∣

∣

∣

∣

∣

4 − λ −14 −λ

∣

∣

∣

∣

∣

∣

= λ2 − 4λ + 4


λ2 − 4λ + 4 = (λ − 2)2 = 0

Eigenvalues: λ1 = λ2 = 2

200

Eigenvectors:

201

3. A =

5 3−6 −1


det (A − λ I) =

∣

∣

∣

∣

∣

∣

5 − λ 3−6 −1 − λ

∣

∣

∣

∣

∣

∣

= λ2 − 4λ + 13


λ2 − 4λ + 13 = 0

Eigenvalues:

λ1 = 2 + 3i, λ2 = 2 − 3i

202

Eigenvectors:

203

NOTE: If a + b i is an eigenvalue

of A with eigenvector α+β i, then

a−b i is also an eigenvalue of A and

α− β i is a corresponding eigenvec-

tor.

204

4. A =

4 −3 5−1 2 −1−1 3 −2


det (A−λ I) =

∣

∣

∣

∣

∣

∣

∣

∣

4 − λ −3 5−1 2 − λ −1−1 3 −2 − λ

∣

∣

∣

∣

∣

∣

∣

∣

= −λ3 + 4λ2 − λ − 6


λ3−4λ2+λ+6 = (λ−3)(λ−2)(λ+1) = 0.

Eigenvals: λ1 = 3, λ2 = 2, λ3 = −1

205

Eigenvectors:

206

5. A =

4 1 −1

2 5 −2

1 1 2


det (A − λ I) =

∣

∣

∣

∣

∣

∣

∣

∣

4 − λ 1 −12 5 − λ −21 1 2 − λ

∣

∣

∣

∣

∣

∣

∣

∣

= −λ3 + 11λ2 − 39λ + 45


λ3−11λ2+39λ−45 = (λ−3)2(λ−5) = 0.

Eigenvalues: λ1 = λ2 = 3, λ3 = 5

207

Eigenvectors:

208

6. A =

−3 1 −1

−7 5 −1

−6 6 −2


det (A−λ I) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

−3 − λ 1 −1

−7 5 − λ −1

−6 6 −2 − λ

∣

∣

∣

∣

∣

∣

∣

∣

∣

= −λ3 + 12λ + 16


λ3−12λ−16 = (λ+2)2(λ−4) = 0.

Eigenvalues: λ1 = λ2 = −2, λ3 = 4

209

Eigenvectors:

210

Some Facts

THEOREM If

v1, v2, . . . , vk

are eigenvectors of a matrix A cor-

responding to distinct eigenvalues

λ1, λ2, . . . , λk,

then v1, v2, . . . , vk are linearly

independent.

211

THEOREM Let A be a (real)

n × n matrix. If the complex num-

ber λ = a + bi is an eigenvalue

of A with corresponding (complex)

eigenvector u+iv, then λ = a−bi,

the conjugate of a + bi, is also an

eigenvalue of A and u − iv is a

corresponding eigenvector.

212

THEOREM Let A be an n × n

matrix with eigenvalues λ1, λ2, . . . , λn.

Then

detA = (−1)nλ1 · λ2 · λ3 · · · · λn.

That is, detA is the product of the

eigenvalues of A.

(The λ’s are not necessarily distinct,

the multiplicity of an eigenvalue may

be greater than 1, and they are not

necessarily real.)

213

Equivalences: (A an n×n matrix)

1. Ax = b has a unique solution.

2. The reduced row echelon form

of A is In.

3. The rank of A is n.

4. A has an inverse.

5. det A 6= 0.

6. 0 is not an eigenvalue of A214

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