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??????13 : Solutions of Linear Systems

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Title: ??????13 : Solutions of Linear Systems


1
??????13 Solutions of Linear Systems
  • ??????? (Kuang-Chi Chen)
  • chichen6_at_mail.tcu.edu.tw

2
Linear Equations and Matrices Solutions of
Linear Systems of Equations
3
Solutions of Linear Systems of Equations
  • 1.6 Solutions of Linear Systems of Equations

4
Row Echelon Form
  • Definition Row echelon form (r.e.f.)
  • An m?n matrix A is said to be in row echelon form
    if
  • (a) All zero rows, if there are any, appear at
    the bottom of the matrix
  • (b) The first nonzero entry from the left of a
    nonzero row is a 1 a leading one of the row
  • (c) For each nonzero row, the leading one appears
    to the right and below any leading ones in
    preceding rows

5
Reduced Row Echelon Form
  • Definition Reduced row echelon form
  • An m?n matrix A is said to be in reduced row
    echelon form if
  • (a) A is in row echelon form
  • (b) If a column contains a leading one, then all
    other entries in that column are zero
  • (???? ???????)

6
Example 1 - in row echelon form
  • E.g. 1

7
Example 2 reduced row echelon form
  • E.g. 2

8
E.g. 2 not reduced row echelon form
  • E.g. 2

,
,
,
Nonzero element above leading 1 in row 2
9
Three Basic Types of Elementary Row Operations
  • Type 1 Interchange
  • row i and row j are interchanged
  • Type 2 Multiply
  • row i row i times c
  • Type 3 Add
  • Add d times row r of A to row s of A
  • row s row s d ? row r

10
Example 3
  • E.g. 3

E1 ?
? E2
E3
11
Row Equivalence
  • Definition Row Equivalence
  • An m?n matrix A is said to be row equivalence
    to an m?n matrix B if B can be obtained by
    applying a finite sequence of elementary row
    operations to the matrix A .

12
Example 4
  • E.g. 4

E3 ?
E1
E2 ?
13
Theorem 1.5
  • Theorem 1.5
  • Every m?n matrix is row equivalent to a matrix
    in row echelon form .

14
E.g. 5 - Procedure of Row Echelon Form
  • E.g. 5
  • Step 1 Find the pivotal column
  • Step 2 Identify the pivot in the pivotal column

15
(contd)
  • E.g. 5
  • Step 3 Interchange if necessary so that the
    pivot is in the 1st row
  • Step 4 Multiply so that the pivot equals to 1

pivot
16
(contd)
  • E.g. 5
  • Step 5 Make all entries in the pivot column,
    except the entry where the pivot was located,
    equal to zero

17
(contd)
  • E.g. 5
  • Step 6 Ignore the first row and repeat
  • ?
  • ? ?

18
Example 6
  • Example 6

19
Remark
  • Remark
  • - There may be more than one matrix in row
    echelon form that is row equivalent to a given
    matrix.
  • - A matrix in row echelon form (r.e.f.) that is
    row equivalent to A is called
  • a row echelon form of A.

20
Theorem 1.6
  • Theorem 1.6
  • - Every m?n matrix is row equivalent to a
    unique matrix in reduced row echelon form.

21
Example 7 r.e.f. to reduced r.e.f.
  • E.g. 7

22
Theorem 1.7
  • Theorem 1.7
  • Let Ax b and Cx d be two linear systems
    each of m equations in n unknowns. If the
    augmented matrices Ab and Cd of these
    systems are row equivalent, then both linear
    systems have the same solutions.

23
Corollary 1.1
  • Corollary 1.1
  • If A and C are row equivalent m?n matrices,
    then the linear system Ax 0 and Cx 0 have
    exactly the same solutions.

24
Gauss-Jordan Reduction Procedure
  • The Gauss-Jordan reduction procedure
  • Step 1. Form the augmented matrix Ab
  • Step 2. Obtain the reduced row echelon form
    Cd of the augmented matrix Ab
    by using elementary row operations
  • Step 3. For each nonzero row of Cd, solve the
    corresponding equation.
  • (augmented matrix ????)

25
Gauss Elimination Procedure
  • The Gauss elimination procedure
  • Step 1. Form the augmented matrix Ab
  • Step 2. Obtain a row echelon form Cd of
    the augmented matrix Ab by using
    elementary row operations
  • Step 3. Solving the linear system corresponding
    to Cd, by back substitution (????).

26
Example 8
  • E.g. 8
  • Solve the linear system by Gauss-Jordan
    reduction
  • - Step 1

27
(contd)
  • E.g. 8 - Solve the linear system by Gauss-Jordan
    reduction
  • - Step 2

28
(contd)
  • E.g. 8 - Solve the linear system by Gauss-Jordan
    reduction
  • - Step 3 x 2
  • y -1
  • z 3

29
Example 9
  • Example 9
  • - Solve the linear system by Gauss-Jordan
    reduction
  • x y 2z 5w 3
  • 2x 5y z 9w -3
  • 2x y z 3w -11
  • x 3y 2z 7w -5

30
(contd)
  • Example 9
  • - Step 1
  • - Step 2

31
(contd)
  • E.g. 9 - Step 3

  • leading variables

  • a free variable

32
Example 10
  • Example 10
  • - Solve the linear system by Gauss-Jordan
    reduction
  • x1 2x2 3x4 x5 2
  • x1 2x2 x3 3x4 x5 2x6 3
  • x1 2x2 3x4 2x5 x6 4
  • 3x1 6x2 x3 9x4 4x5 3x6 9

33
(contd)
  • Example 10
  • - Step 1
  • - Step 2

34
(contd)
  • Example 10 - Step 3

  • leading variables

  • free variables

35
Example 11
  • Example 11
  • - Solve the linear system by Gauss elimination
  • x 2y 3z 9
  • 2x y z 8
  • 3x z 3

36
(contd)
  • Example 11
  • - Step 1
  • - Step 2

37
(contd)
  • Example 11 - Step 3
  • - By back substitution

38
Example 12
  • Example 12
  • - Solve the linear system by Gauss elimination
  • x 2y 3z 4w 5
  • x 3y 5z 7w 11
  • x z w -6

39
(contd)
  • Example 12
  • - Step 1
  • - Step 2
  • - Step 3 ? 0x 0y 0z 0w 1 ? No
    solutions !!

40
Consistent and Inconsistent
  • Consistent and inconsistent
  • - Consistent Linear systems with at least one
    solution
  • - Inconsistent Linear systems with no solutions

41
Homogeneous Systems
  • A system of linear equations is said to be
    homogeneous if all the constant terms are zeros.
  • a11x1 a12x2 a1nxn 0
  • a21x1 a22x2 a2nxn 0
  • am1x1 am2x2 amnxn 0
  • ? Ax 0
  • Thus, a homogeneous system always has the
    solution x1 x2 xn 0 ? the trivial
    solution

42
Example 13
  • Example 13

43
Example 14
  • Example 14

44
Theorem 1.8
  • Theorem 1.8
  • A homogeneous system of m equations in n
    unknowns has a non-trivial solution if m lt n,
    that is, if the number of unknowns exceeds the
    number of equations.
  • namely, a homogeneous system has more
    variables than equations has many solutions.
  • (a homogeneous system???? non-trivial
    solution ???)

45
Example 15 - A Homogeneous System
  • E.g. 15

46
(contd)
  • If let

47
A Homogeneous System Example
x xp xh xp is a particular solution
to the given system Axp b , where b
3 -3 -11 -5T xh is a solution to the
associated homogeneous system Axh 0 .
48
Polynomial Interpolation
  • Polynomial Interpolation
  • - The general form
  • y an 1xn 1 an 2xn 2 a1x
    a0
  • E.g. n 3, y a2x2 a1x a0
  • Given three distinct points (x1 , y1), (x2 , y2),
    (x3 , y3),
  • we have
  • y1 a2x12 a1x1 a0
  • y2 a2x22 a1x2 a0
  • y3 a2x32 a1x3 a0

49
Example 16
  • Example 16 - Find the quadratic polynomial that
    interpolates the points (1, 3), (2, 4), (3, 7)

50
Example 17 Temperature Distribution
  • T1 (260 100 T2 T3 )/4 or 4T1 T2
    T3 160
  • T2 (T1 100 40 T4 )/4 or -T1 4T2
    T4 140
  • T3 (60 T1 T4 0)/4 or -T1 4T3
    T4 60
  • T4 (T2 T3 40 0)/4 or -T2 T3
    4T4 40
  • ?
  • ? T1 65, T2 60, T3 40, T4 35 .

51
Linear Equations and Matrices The Inverse of A
Matrix
52
The Inverse of A Matrix
  • 1.7 The inverse of a matrix
  • Definition
  • - An n?n matrix A is called nonsingular (or
    invertible ???) if there exists an n?n matrix B
    such that AB BA In .
  • - The matrix B is called the inverse of A
  • - If there exists no such matrix B, then A is
    called singular (or noninvertible)
  • - A is also an inverse of B

53
Example 1
  • Example 1
  • ? AB BA I2
  • - B is an inverse of A and A is nonsingular.

54
Theorem 1.9
  • Theorem 1.9
  • An inverse of a matrix, if exists, is unique.
  • (proof)
  • Let B and C be inverses of A.
  • Then AB BA In, and AC CA In.
  • Thus, C(AB) CIn
  • (CA)B C
  • InB C , i.e., B C .

55
Example 2 - Find the Inverse
For the matrix A, find the inverse If exists,
let the inverse A-1 be such that
56
(contd)
and
57
Example 3
  • Example 3

and
? No solution singular
58
Theorem 1.10
  • Thm. 1.10 - Properties of an inverse
  • - If A is nonsingular, then A-1 is nonsingular
  • and (A-1)-1 A
  • - If A and B are nonsingular matrices, then AB
    is nonsingular and (AB)-1 B-1 A-1
  • - If A is a nonsingular matrix, then (AT)-1
    (A-1)T .

59
Example 4
  • Example 4

? and
60
Corollary1.2
  • Corollary 1.2
  • - If A1 , A2 , , Ar are n?n nonsingular
    matrices, then (A1 A2 Ar) is nonsingular and
    (A1 A2 Ar)-1 Ar-1 A2-1 A1-1 .

61
Theorem 1.11
  • Theorem 1.11
  • Suppose that A, B are n?n matrices,
  • - If AB In , then BA In
  • - If BA In , then AB In .

62
The Way to Find A-1

  • A practical method for finding A-1
  • Step 1. Form the 2?2n matrix A In obtained by
    adjoining the identity matrix In to the given
    matrix A
  • Step 2. Compute the reduced row echelon form of
    the matrix obtained in Step 1 by using elementary
    row operations
  • Step 3. Suppose that Step 2 has produced the
    matrix C D in reduced row echelon form
  • If C In , then D A-1
  • If C ? In , then C has a row of zeros and the
    matrix A is singular .

63
Example 5 Find the Inverse
  • E.g. 5 Find the inverse

64
Example 6 Find the Inverse
  • E.g. 6 - Find the inverse

The left-half matrix cannot have a one in the (3,
3) location, the reduced echelon form cannot be
I3. Thus A-1 does not exist.
65
Theorem 1.12 1.13
  • Theorem 1.12
  • An n?n matrix is nonsingular iff it is row
    equivalence to In .
  • Theorem 1.13
  • If A is an n?n matrix, the homogeneous system
    Ax 0 has a nontrivial solution iff
  • A is singular.

66
Proof of Theorem 1.13
  • Proof of Theorem 1.13
  • Suppose that A is nonsingular, then A-1 exists
    and
  • A-1(Ax) A-1 0
  • (A-1A)x 0
  • In x 0
  • x 0 ? Ax 0 has a trivial
    solution
  • (contradiction to a non-trivial solution,
    hence A must be singular)

67
Example 8
  • Example 8
  • Consider the homogeneous system Ax 0, where
    A is the matrix (A is nonsingular)

Gauss-Jordan reduction
The trivial solution x 0
68
Example 9
  • Example 9
  • - Consider the homogeneous system Ax 0, where
    A is the matrix (A is singular)

69
Theorem 1.14
  • Theorem 1.14
  • If A is an n?n matrix, then A is nonsingular
    iff the linear system Ax b has a unique
    solution for every n?1 matrix b .

70
Summary The Symmetry, Singularity, Inverse of A
Matrix
71
Some Special Matrix
  • 4. A square matrix A is said to be antisymmetric
    if -AT A. (i) If A is square, prove that A AT
    is symmetric and A AT is antisymmetric
  • (ii) any square matrix A can be decomposed
    into the sum of a symmetric matrix B and an
    antisymmetric matrix C A B C .
  • 5. Given two symmetric matrices of the same
    size, A and B, then a necessary and sufficient
    condition for the product AB to be symmetric is
    that AB BA.

72
Some Special Matrix
  • 1. A square matrix A is said to be normal if AAT
    ATA. All symmetric matrices are normal
  • 2. A square matrix A is said to be idempotent if
    A2 A. If A is idempotent then AT is also
    idempotent
  • 3. A square matrix A is said to nilpotent if
    there is a positive integer p such that Ap O.
    The least integer such that Ap O is called the
    degree of nilpotency of the matrix. If A is
    nilpotent, then AT is also nilpotent with the
    same degree of nilpotency.

73
List of Nonsingular Equivalences
  • Nonsingular equivalences
  • 1. A is nonsingular
  • 2. x 0 is the only solution to Ax 0
  • 3. A is row equivalence to In
  • 4. The linear system Ax b has a unique solution
    for every n?1matrix b .

74
Properties of Matrix Inverse
  • Properties of Matrix Inverse
  • 1. (A-1)-1 A
  • 2. (cA)-1 (1/c)A-1 , where c is a nonzero
    scalar
  • 3. (AB)-1 B-1A-1
  • 4. (An)-1 (A-1)n
  • 5. (AT)-1 (A-1)T , where T transpose.

75
Conditions of Matrix Inverse
  • A matrix has no inverse, if
  • (i) two rows are equal
  • (ii) two columns are equal (Use the transpose)
  • (iii) it has a column of zeros.

76
The Inverse of 2?2 Matrix
  • If A , show that A-1
    .
  • Note The cancellation law doesnt hold.
  • That is, AB AC doesnt imply that B C .
  • Also, AB O doesnt imply that A O or B O.
  • However, if A is an invertible matrix, then
  • if AB AC , then B C
  • if AB 0, then B 0 .
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