1.1 Systems of Linear Equations Basic concept linear equation(
) system of linear equations( ), and its solution Matrix( )
Slide 3
1.1.1 what is a linear equation? Definition 1 (linear equation(
)). A linear equation in the variables x 1,,x n is an equation of
the form a 1 x 1 + a 2 x 2 +... + a n x n = b (1) where b and the
coefficients a 1, ,a n are real or complex numbers. eg.
Slide 4
What Is System of Linear Equations? Definition 2 (system of
linear equations( )). A system of linear equations (or linear
system) is a collection of one or more linear equations involving
the same variables- x 1, , x n a 1,1 x 1 + a 1,2 x 2 +... + a 1,n x
n = b 1 a 2,1 x 1 + a 2,2 x 2 +... + a 2,n x n = b 2 2 ... a m,1 x
1 + a m,2 x 2 +... + a m,n x n = b m
Slide 5
Slide 6
1.1.2 Solution of System of Linear Equations Definition 3
(solution())). A list (S 1, S 2, S n ) of numbers is called a
solution of (2) iff (i.e. if and only if) all the equations in (2)
are satisfied by substituting S 1, S 2, S n for X 1, X 2, X n. The
set of all solutions of (2) is called the solution set () of (2).
Two systems of linear equations are said to be equivalent () if
they have the same solution set.
Slide 7
1.1.2 Solution of System of Linear Equations A system of linear
equations has either 1. No solution, or 2. Exactly one solution, or
3. Infinitely many solutions. Definition 4 (consistence ( )). A
system of linear equations is said to be consistent if its solution
set is nonempty (i.e. either one solution or infinitely many
solutions), otherwise it is inconsistent. consistent
inconsistent
Slide 8
1.1.2 Solution of System of Linear Equations Fig(a). Exactly
one solution Fig(b). no solution Fig(c). Infinitely many
solutions
Slide 9
1.1.3 Matrix Notation P.4 Matrix Notation Coefficient matrix
augmented matrix The size of a Matrix: how many rows and columns it
has.
Slide 10
1.1.3 Matrix Notation Definition 5 (matrix ( )). A table of
numbers with m rows ( ) and n columns ( ) as above is called an m n
matrix. we normally use a capital letter such as A, B, X etc. to
denote a matrix. Coefficient matrixaugmented matrix
Slide 11
1.1.4 Solving a Linear System P.5 Basic strategy ( ). To
replace one system with an equivalent system (one with the same
solution set) that is easier to solve Three basic operations (
elementary operation( ) to simplify a linear system 1. replace( )
one equation by the sum of itself and a multiple of another
equation 2. interchange( ) two equations 3. Scaling( ) all the
terms in an equation by a nonzero constant
Slide 12
Solving a Linear System 4*[eq.1]+[eq.3] ()*[eq.2]
3*[eq.2]+[eq.3] 4*[eq.3]+[eq.2] -1*[eq.3]+[eq.1] Sol: Upper
triangular Back subsitution
Slide 13
Sol: augmented matrix
Slide 14
Slide 15
Definition 6 (Row Equivalence ( )). If matrix A can be
transformed into matrix B by applying a series of elementary row
operations on A, then we say A is row equivalent to B and denote
this equivalence by A ~ B. (P.7) If the augmented matrices of two
linear systems are row equivalent, then the two systems have the
same solution set. (P.8)
Slide 16
1.1.5 Existence and Uniqueness Questions P.8 Two fundamental
questions about a linear system 1. Is the system consistent; that
is, does at least one solution exist? 2. If a solution exists, is
it the only one; that is, is the solution unique?
Slide 17
Eg: Determine if the following system is consistent Sol: From
example1, we have We know x 3, and substitute the value of x 3 into
eq.2 could get x 2, then could determine x 1 from eq.1. So a
solution exists; the system is consistent.
Slide 18
Eg:Determine if the following system is consistent: Sol: The
equation 0x 1 +0x 2 +0x 3 =(5/2) is never true, so the system is
inconsistent.
Slide 19
1.2 Row Reduction and Echelon Forms P.14 Basic concept: leading
entry ( ) (row) echelon form ( ) echelon matrix ( ) reduced (row)
echelon form ( ), reduced (row) echelon matrix( ) pivot position (
)
Slide 20
1.2.1 Echelon Forms P.14 *Definition 1:leading entry( ): the
first nonzero entry in a nonzero row. Definition 2 A rectangular
matrix is in echelon form (or row echelon form) if : 1. All nonzero
rows are above any rows of all zeros. 2. Each leading entry of a
row is in a column to the right of the leading entry of the row
above it. 3. All entries in a column below a leading entry are
zeros.
Slide 21
The following matrices are in echelon form(upper triangular
matrix):
Slide 22
reduced echelon form (or row reduced echelon form) Definition 3
A rectangular matrix is in reduced echelon form (or row reduced
echelon form,RREF) if : 1. All nonzero rows are above any rows of
all zeros. 2. Each leading entry of a row is in a column to the
right of the leading entry of the row above it. 3. All entries in a
column below a leading entry are zeros. 4. The leading entry in
each nonzero row is 1. 5. Each leading 1 is the only nonzero entry
in its column.
Slide 23
The following matrices are in reduced echelon form:
Slide 24
Theorem 1 : Uniqueness of the Reduced Echelon Form (p.15) Each
matrix is row equivalent to one and only one reduced echelon
matrix. Each matrix is row equivalent to one and only one reduced
echelon matrix. If a matrix A is row equivalent to an echelon
matrix U, we call U an echelon form of A; If U is in reduced
echelon form, we call U the reduced echelon form of A.
Slide 25
1.2.2 Pivot position( ) P.15 pivot: A pivot in a row echelon
matrix U is a leading nonzero entry in a nonzero row. Definition 4
pivot position: a position of a leading entry in an echelon form of
the matrix. (P.16) pivot column: a column that contains a pivot
position.
Slide 26
Sol: Interchange row1 and row4 Adding multiples of the first
rows below: Example 2: Row reduce the matrix A below to echelon
form, and locate the pivot columns of A.
Slide 27
Adding -5/2 times row 2 to row3, and add 3/2 times row 2 to row
4 interchange rows 3 and 4 Note There is no more than one in any
row. There is no more than one in any colomn.
Slide 28
1.2.3 The Row Reduction Algorithm( ) P.17 Why? The reduced
echelon form of a matrix A has the same solution as the original
one. More, the reduced echelon form is easy for computing. Step1
Begin with the leftmost nonzero column. Step2 Select a nonzero
entry in the pivot column as a pivot. Step3 Use row replacement
operations to create zeros in all positions below the pivot. Step4
Apply steps 1-3 to the submatrix that remains. Repeat the process
until there are no more nonzero rows to modify. Step5 Beginning
with the rightmost pivot and working upward and to the left, create
zeros above each pivot
Slide 29
Example 3: Transform the following matrix into reduced echelon:
Sol: Step1:Step2: Step3:
Slide 30
Step4: Step5: (1) (2) (3) (4) The combination of steps 1-4 is
called the forward phase of the row reductions algorithm. Steps 5
is called backward phase. (1) (2)
Slide 31
1.2.4. Solution of Linear Systems P.20 augmented
matrixAssociated system of equation solution Basic variable( ): any
variable that corresponds to a pivot column in the augmented matrix
of a system. free variable( ) all nonbasic variables. (5) (4)
Slide 32
Example 4: Find the general solution( ) of the following linear
system Sol:
Slide 33
The associated system now is The general solution is: (7)
Slide 34
1.2.5 Parametric Descriptions of Solution Sets P.22 Solving a
system amounts to finding a parametric description of the solution
set or determine that the solution set is empty. The solution has
many parametric descriptions. We make the arbitrary convention of
always using the free variables as the parameters for describing a
solution set.
Slide 35
1.2.5 Parametric Descriptions of Solution Sets P.22
Back-Substitution A computer program would solve system by
back-substitution
Slide 36
1.2.6 Existence and Uniqueness Questions P.23 (8)
Slide 37
Slide 38
Existence and Uniqueness Questions Theorem 2 Existence and
Uniqueness Theorem A linear system is consistent if and only if the
rightmost column of the augmented matrix is not a pivot column that
is, if and only if an echelon form of the augmented matrix has no
row of the form A linear system is consistent if and only if the
rightmost column of the augmented matrix is not a pivot column that
is, if and only if an echelon form of the augmented matrix has no
row of the form If a linear system is consistent, then the solution
set contains either (i) a unique solution, when there are no free
variable, or (ii) infinitely many solutions, when there is at least
one free variable If a linear system is consistent, then the
solution set contains either (i) a unique solution, when there are
no free variable, or (ii) infinitely many solutions, when there is
at least one free variable
Slide 39
Solutions of Linear Systems( )
Slide 40
Slide 41
Slide 42
Using Row Reduction to Solve A Linear System 1: Write the
augmented matrix of the system. 2: Use the row reduction algorithm
to obtain an equivalent augmented matrix in echelon form. If the
system is inconsistent, Stop. 3: Continue row reduction to obtain
the reduced echelon form. 4: Write the system of equations
corresponding to the matrix obtained in step3. 5: Rewrite each
nonzero equation form step4 so that its one basic variable is
expressed in terms of any free variables appearing in the
equation.
1.3 Vector Equations P.28 Vectors in R 2 Geometric Description
of R 2 Vectors in R 3 Vectors in R n Linear Combination A Geometric
Description of Span{v} and Span{u,v} Linear Combinations in
Applications
Slide 45
1.3.1 Vector P.28 Definition 1 (vectors, ) A matrix with only
one column is called a column vector, or simply a vector. 1.3.1
Vectors in R 2 A two-dimensional vector is a pair of numbers,
surrounded by brackets( ).
Slide 46
1.3 Vector Equations Vectors in R 2 Notation: Different people
use different notation for vector. v (boldface), (use arrows)
Slide 47
Vectors in R 2
Slide 48
vectors are equal: If and only if they have the same
corresponding entries. eg: Vector Addition: We add vectors in the
obvious way, componentwise =
Slide 49
Scalar Multiplication( ) : Notes: the vector cv has the same
direction as v if c > 0,and the direction opposite to v if c
< 0. Geometric Description of R 2 Vector as points Vectors with
arrows
Slide 50
Slide 51
Parallelogram Rule ( )For Addition If u and v in are
represented as points in the plane, then u+v corresponds to the
fourth vertex of the parallelogram whose other vertices are u,0 and
v. Fig. The parallelogram rule
Slide 52
1.3.2 Vectors in R 3 - vectors in R 3 are 31 column matrices
with three entries. - represented geometrically by points in a 3D
coordinate space Fig. Scalar multiples in R 3
Slide 53
1.3.3 Vectors in R n (n ) R n denotes the collection of all
lists of n real numbers - written as n1 column matrices - zero
vector
Slide 54
Algebraic Properties( ) of R n For all u, v, w in R n and all
scalars c and d: 0
Slide 55
1.3.4. Linear Combination( ) p. 32 Definition 4(Linear
combination):Given vectors v 1, v 2,,v p in R n and given scalars c
1, c 2,,c p, the vector y defined by is called a linear Combination
of v 1, v 2,,v p with weight ( ) c 1, c 2,,c p. eg. ?
Slide 56
Slide 57
Example
Slide 58
Let and Determine whether b can be generated as a linear
combination of a 1 and a 2. ( b a 1, a 2 ) That is, determine
whether x 1 and x 2 exist such that x 1 a 1 + x 2 a 2 = b. Sol.
and
Slide 59
Get the system: Solve the system: so:Hence b is a linear
combination of a 1 and a 2
Slide 60
Facts A vector equation has the same solution set as the linear
system whose augmented matrix is In particular, b can be generated
by a linear combination of a 1, ,a n if and only if there exists a
solution to the linear system corresponding to
Slide 61
_
Slide 62
1.3.5 Span ( ) P.35 b Span{ v 1, v 2,, v n }, x 1 v 1 +x 2 v 2
+..+ x n v n =b
Slide 63
A Geometric Description of Span{v} and Span{u,v}
Slide 64
Slide 65
Slide 66
Slide 67
Slide 68
Eg: A company manufactures two products. For $1.00 worth of
product B, the company spend $.45 on materials, $.25 on labor, and
$.15 on overhead. For $1.00 worth of product C, the company spend
$.40 on materials, $.30 on labor, and $.30 on overhead. Let A.what
economic interpretation can be given to the vector 100b? B. Suppose
the company wishes to manufacture x1 dollars worth of product B and
x2 dollars worth of product C. Give a vector that describes the
various costs the company will have.
Slide 69
So l. A. The vector 100b list the various costs for producing
$100 worth of product B, $45 for material, $25 for labor, and $15
for overhead. B. The costs of manufacturing x 1 dollars worth of B
are given by the vector x 1 b, and the costs of manufacturing x 2
dollars worth of C are given by the vector x 2 c. Hence the total
costs for both products by the vector x 1 b+x 2 c