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MATH 233 - Linear Algebra I
Lecture Notes
Cesar O
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1 What is a system of linear equations?
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2 Matrices
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3 Solving linear systems
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4 Geometric interpretation of the solution set
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2 Row Reduction and Echelon Forms
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3 Vector Equations
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4 The
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Matrix-vector multiplication and linear combinations
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5 Homogeneous and Nonhomogeneous Systems
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6 Linear Independence
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7 Introduction to Linear Mappings
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3
CONTENTS
8 Onto, One-to-One, and Standard Matrix
8
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8
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8
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67
67
69
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9 Matrix Algebra
9
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9
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9
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75
75
76
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10 Invertible Matrices
10
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10
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10
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83
83
85
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11 Determinants
11
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11
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11
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89
89
93
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12 Properties of the Determinant
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12
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97
12
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100
12
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13 Applications of the Determinant
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13
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103
13
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106
13
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14 Vector Spaces
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14
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109
14
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15 Linear Maps
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15
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117
15
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121
16 Linear Independence, Bases, and
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16
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16
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Dimension
125
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126
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1 The Rank of a Matrix
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1 Coordinates
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2 Coordinate Mappings
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3 Matrix Representation of a Linear Map
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1 Review of Coordinate Mappings on Rn
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2 Change of Basis
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1 Inner Product on Rn
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2 Orthogonality
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3 Coordinates in an Orthonormal Basis
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1 Eigenvectors and Eigenvalues
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2 When λ = 0 is an eigenvalue
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1 The Characteristic Polynomial of a Matrix
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2 Eigenvalues and Similarity Transformations
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1 Eigenvalues of Triangular Matrices
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2 Diagonalization
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3 Conditions for Diagonalization
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1 Symmetric Matrices
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2 Eigenvectors of Symmetric Matrices
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3 Symmetric Matrices are Diagonalizable
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1
25
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3
187
187
188
188
PageRank Algortihm
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Search Engine Retrieval Process
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192
Computation of the PageRank Vector
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1 Discrete Dynamical Systems
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2 Population Model
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3 Stability of Discrete Dynamical Systems
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We will introduce matrices as a convenient structure to represent and solve linear
systems
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1
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1: A system of m linear equations in n unknown variables x1 , x2 ,
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1)
= bm
The numbers aij are called the coefficients of the linear system; because there are m equations and n unknown variables there are thefore m × n coefficients
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, sn ) that satisfy the system of linear equations (1
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In other words, if we substitute the list of numbers (s1 , s2 ,
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, xn ) in equation (1
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We call such a list (s1 , s2 ,
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Notice
that we say “a solution” because there may be more than one
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As an example of a linear system, below is a linear
1
Systems of Linear Equations
system consisting of m = 2 equations and n = 3 unknowns:
x1 − 5x2 − 7x3 = 0
5x2 + 11x3 = 1
Here is a linear system consisting of m = 3 equations and n = 2 unknowns:
−5x1 + x2 = −1
πx1 − 5x2 = 0
√
63x1 − 2x2 = −7
And finally, below is a linear system consisting of m = 4 equations and n = 6 unknowns:
−5x1 + x3 − 44x4 − 55x6
√
πx1 − 5x2 − x3 + 4x4 − 5x5 + 5x6
√
1
63x1 − 2x2 − 15 x3 + ln(3)x4 + 4x5 − 33
x6
√
1
63x1 − 2x2 − x3 − 18 x4 − 5x6
5
= −1
=0
=0
=5
Example 1
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Verify that (1, 2, −4) is a solution to the system of equations
2x1 + 2x2 + x3 = 2
x1 + 3x2 − x3 = 11
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The number of equations is m = 2 and the number of unknowns is n = 3
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And
b1 = 0 and b2 = 11
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Thus, (1, −1, 2) is not a solution to the system
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If this is the case, we say that the linear
system is inconsistent:
2
Lecture 1
INCONSISTENT ⇔ NO SOLUTION
A linear system is called consistent if it has at least one solution:
CONSISTENT ⇔ AT LEAST ONE SOLUTION
We will see shortly that a consistent linear system will have either just one solution or
infinitely many solutions
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If
it has multiple solutions, then it will have infinitely many solutions
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3
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−x1 + x2 = 3
x1 − x2 = 1
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If we add the two equations we get
0=4
which is a contradiction
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Example 1
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Let t be an arbitrary real number and let
s1 = − 32 − 2t
s2 = 23 + t
s3 = t
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Solution
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Because we can vary t arbitrarily, we get an
infinite number of solutions parameterized by t
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3
Systems of Linear Equations
1
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Informally speaking, a matrix is an
array or table consisting of rows and columns
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In general, a matrix with m rows and
n columns is a m × n matrix and the set of all such matrices will be denoted by Mm×n
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The entry of A in the ith row and jth column will be
denoted by aij
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For example, here is a row vector
u = 1 −3 4
and here is a column vector
3
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For example, for the linear system
5x1 − 3x2 + 8x3 = −1
x1 + 4x2 − 6x3 = 0
2x2 + 4x3 = 3
the coefficient matrix
5
A= 1
0
A, the output vector b, and the augmented matrix [A b] are:
−3 8
−1
5 −3 8 −1
4 −6 , b = 0 , [A b] = 1 4 −6 0
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Using our previously defined notation,
we can write this as A ∈ Mm×n
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For example, the linear system associated to the augmented matrix
1 4 −2 8 12
0 1 −7 2 −4
0 0 5 −1 7
is
x1 + 4x2 − 2x3 + 8x4 = 12
x2 − 7x3 + 2x4 = −4
5x3 − x4 = 7
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Matrix algebra is a fascinating subject with numerous
applications in every branch of engineering, medicine, statistics, mathematics, finance, biology, chemistry, etc
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3
Solving linear systems
In algebra, you learned to solve equations by first “simplifying” them using operations that
do not alter the solution set
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We can do
similar operations on a linear system
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Interchange two equations
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Multiply an equation by a nonzero constant
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Add a multiple of one equation to another
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The idea is to apply these operations iteratively to simplify the linear system to a point where one can easily write down the solution
set
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In this case, we call the operations elementary row operations,
and the process of simplifying the linear system using these operations is called row reduction
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This is
best explained via an example
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5
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the augmented matrix
0 −2 −4
1 −1 0
0 1
1
Solution
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The third row
corresponds to the equation x3 = 1
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The first row corresponds to the equation
x1 − 2x3 = −4
and therefore
x1 = −4 + 2x3 = −4 + 2 = −2
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5
Systems of Linear Equations
Example 1
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Solve the linear system using elementary row operations
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Our goal is to perform elementary row operations to obtain a triangular structure
and then use back substitution to solve
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2 −3 4 −3
Interchange Row 1 (R1 ) and Row 2
−3 2
4
1
0 −2
2 −3 4
(R2 ):
12
1
0 −2 −4
R ↔R2
−3 2
−4 −−1−−→
4 12
−3
2 −3 4 −3
As you will see, this first operation will simplify the next step
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So now use back substitution to solve
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From the second equation we obtain that x2 − x3 = 0,
and thus x2 = 1
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Thus, the solution
to the original system is (−2, 1, 1)
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The original augmented matrix of the previous example is
−3 2
4 12
−3x1 + 2x2 + 4x3 = 12
→
0 −2 −4
M= 1
x1 − 2x3 = −4
2 −3 4 −3
2x1 − 3x2 + 4x3 = −3
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Although the two augmented matrices M and N are clearly distinct, it is a fact that they
have the same solution set
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7
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x1 + 2x3 = 1
x2 + x3 = 0
2x1 + 4x3 = 1
Solution
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e
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In general, if we obtain a row in an
augmented matrix of the form
0 0 0 ··· 0 c
where c is a nonzero number, then the linear system is inconsistent
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However, a row of the form
0 1 0 0 0
corresponds to the equation x2 = 0 which is perfectly valid
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4
Geometric interpretation of the solution set
The set of points (x1 , x2 ) that satisfy the linear system
x1 − 2x2 = −1
−x1 + 3x2 = 3
(1
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The solution
for this system is (3, 2)
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1
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1: The intersection point of the two lines is the solution of the linear system (1
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3)
8
Lecture 1
is the intersection of the three planes determined by the equations of the system
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In the case of a consistent system of two equations,
the solution set is the line of intersection of the two planes determined by the equations of
the system, see Figure 1
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−4x1 + 5x2 + 9x3 = −9
x1 − 2x2 + x3 = 0
the solution set is this line
Figure 1
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3)
After this lecture you should know the following:
• what a linear system is
• what it means for a linear system to be consistent and inconsistent
• what matrices are
• what are the matrices associated to a linear system
• what the elementary row operations are and how to apply them to simplify a linear
system
• what it means for two matrices to be row equivalent
• how to use the method of back substitution to solve a linear system
• what an inconsistent row is
• how to identify using elementary row operations when a linear system is inconsistent
• the geometric interpretation of the solution set of a linear system
9
Systems of Linear Equations
10
Lecture 2
Lecture 2
Row Reduction and Echelon Forms
In this lecture, we will get more practice with row reduction and in the process introduce
two important types of matrix forms
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Lastly, we will introduce a convenient
parameter called the rank of a matrix
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1
Row echelon form (REF)
Consider the linear system
x1 + 5x2 − 2x4 − x5 + 7x6 = −4
2x2 − 2x3 + 3x6 = 0
−9x4 − x5 + x6 = −1
5x5 + x6 = 5
0=0
having augmented matrix
1
0
0
0
0
5 0 −2 −1 7 −4
2 −2 0
0 3 0
0 0 −9 −1 1 −1
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All nonzero rows are above any rows of all zeros
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The leftmost nonzero entry of a row is to the right of the leftmost nonzero entry of
the row above it
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In
REF, the leftmost nonzero entry in a row is called a leading entry:
1 5 0 −2 −1 7 −4
0 2 −2 0
0 3 0
0 0 0 −9 −1 1 −1
0 0 0
0
5 1 5
0 0 0
0
0 0 0
A consequence of property P2 is that every entry below a leading entry is zero:
1 5 0 −2 −4 −1 −7
0 2 −2 0
0
3
0
0 0 0 −9 −1 1 −1
0 0 0
0
5
1
5
0 0 0
0
0
0
0
We can perform elementary row operations, or row reduction, to transform a matrix into
REF
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1
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Use elementary row
operations to put them in REF
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Matrix M fails property P1
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To put N in REF we
R3 :
7 5 0 −3
7
−2R2 +R3
0 3 −1 1 −
−−−−→ 0
0 6 −5 2
0
perform the operation −2R2 + R3 →
5 0 −3
3 −1 1
0 −3 0
Why is REF useful? Certain properties of a matrix can be easily deduced if it is in REF
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If an augmented matrix
is in REF, we can use back substitution to solve the system, just as we did in Lecture 1
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Substituting x3 = 2 into
the second equation we obtain that x2 = 3
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2
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A matrix is in RREF if it is in REF (so it satisfies properties P1 and P2) and in
addition satisfies the following properties:
P3
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P4
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A leading 1 in the RREF of a matrix
in RREF:
1
0
0
has three pivots:
is called a pivot
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2
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0 3 −6 6 4 −5
3 −7 8 −5 8 9
3 −9 12 −9 6 15
Solution
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In the
next phase, we work bottom-to-top and create zeros above the leading 1’s
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Example 2
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Use row reduction to solve the linear system
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The augmented matrix is
2 4 6 8
1 2 4 8
3 6 9 12
14
Lecture 2
Create a leading 1 in the first row:
2 4 6 8
1
2
3
4
1
R1
1 2 4 8 −2−→
1 2 4 8
3 6 9 12
3 6 9 12
Create zeros under the first leading
1 2 3
1 2 4
3 6 9
1 2 3
0 0 1
3 6 9
1:
4
1
−R1 +R2
8 −−−−−→ 0
12
3
4
1
−3R +R3
0
4 −−−1−−→
12
0
2 3 4
0 1 4
6 9 12
2 3 4
0 1 4
0 0 0
The system is consistent, however, there are only 2 nonzero rows but 3 unknown variables
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The second row
in the augmented matrix is equivalent to the equation:
x3 = 4
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We now must choose one of the variables x1 or x2 to be a parameter, say t, and solve for the
remaining variable
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We can therefore write the solution set for the linear system as
x1 = −8 − 2t
x2 = t
x3 = 4
(2
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If we had chosen x1 to be the parameter, say x1 = t,
then the solution set can be written as
x1 = t
x2 = −4 − 12 t
x3 = 4
(2
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1) and (2
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15
Row Reduction and Echelon Forms
In general, if a linear system has n unknown variables and the row reduced augmented
matrix has r leading entries, then the number of free parameters d in the solution set is
d = n − r
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In the previous example, there are n = 3 unknown variables and
the row reduced augmented matrix contained r = 2 leading entries
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Because the number of leading entries r in the row reduced coefficient matrix determine the
number of free parameters, we will refer to r as the rank of the coefficient matrix:
r = rank(A)
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Example 2
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Solve the linear system represented by the augmented matrix
1 −7 2 −5 8 10
0 1 −3 3
1 −5
0 0
0
1 −1 4
Solution
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Thus, the solution set is parameterized by d = 5 − 3 = 2 free variables,
call them t and s
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We choose x5
to be the first parameter so we set x5 = t
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The second equation of
the augmented matrix is
x2 − 3x3 + 3x4 + x5 = −5
and the unassigned variables are x2 and x3
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Then
x2 = −5 + 3x3 − 3x4 − x5
= −5 + 3s − 3(4 + t) − t
= −17 − 4t + 3s
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Choose arbitrary numbers for t and s and
substitute the corresponding list (x1 , x2 ,
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2
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Theorem 2
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One of the following
distinct possibilities will occur:
1
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2
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3
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, the linear system is inconsistent and thus has no solution
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, the linear
system is consistent and has only one (and thus unique) solution
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In Case 3
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This case occurs when r < n
and thus d = n − r > 0 is the number of free parameters
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5)
17
Row Reduction and Echelon Forms
18
Lecture 3
Lecture 3
Vector Equations
In this lecture, we introduce vectors and vector equations
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As we will see, solving
the linear combination problem reduces to solving a linear system of equations
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1
Vectors in Rn
Recall that a column vector in Rn is a n × 1 matrix
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It is important to emphasize that a
vector in Rn is simply a list of n numbers; you are safe (and highly encouraged!) to forget
the idea that a vector is an object with an arrow
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−1
Here is a vector in R3 :
Here is a vector in R6 :
−3
v = 0
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0
3
To indicate that v is a vector in Rn , we will use the notation v ∈ Rn
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When we write vectors within a paragraph, we willwrite
−1
them using list notation instead of column notation, e
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, v = (−1, 4) instead of v =
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For example,
here is the addition of two vectors:
4
4
0
−5 −3 −8
+ =
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5
15
And here are both operations combined:
4
−2
−8
−6
−14
−2 −8 + 3 9 = 16 + 27 = 43
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As the following example illustrates,
vectors can be used in a natural way to represent the solution of a linear system
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1
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The number of unknowns is n = 5 and the associated coefficient matrix A has
rank r = 3
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This
system was considered in Example 2
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The solution in vector form therefore takes the
form
19
−31
−89
−89 − 31t1 + 19t2
x1
3
−4
x2 −17 − 4t1 + 3t2 −17
= 0 + t1 0 + t2 1
t2
x=
x3 =
0
1
4
x4
4 + t1
0
1
0
t1
x5
20