linear algebra (2009 fall) chapter 1 matrices and systems of...
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Linear Algebra
Linear Algebra (2009 Fall)Chapter 1 Matrices and Systems of Equations
Chih-Wei Yi
Dept. of Computer ScienceNational Chiao Tung University
October 29, 2009
Linear Algebra
Systems of Linear Equations
Linear Systems
Linear Systems
Linear Algebra
Systems of Linear Equations
Linear Systems
Linear Equations
De�nition (Linear Equations)
A linear equation in n unknowns (variables) x1, x2, ..., xn:
a1x1 + a2x2 + � � �+ anxn = b.
x1, x2, � � � , xn: variables (x1: leading varialbes)a1, a2, � � � , an: constants and called coe¢ cients (a1: leadingcoe¢ cient)
b: constant term
Example
2x + 3y = 6 is a linear equation in 2 unknowns, but y = sin x andxy = 1 are not linear equations.
Linear Algebra
Systems of Linear Equations
Linear Systems
Examples
Which are linear equations?
2x � 3y = 43x � 4xy = 01x �
2y = 3
x2 + y2 = 1
2 sin x + y = 4
(sin 2) x + y = 10
2x1 + e3x2 = log 5
x2 + 3x + 2 = 0
Linear Algebra
Systems of Linear Equations
Linear Systems
Solutions of Linear Equations
De�nition (Solutions)
Assume s1, s2, . . . , sn are n real numbers.x1 = s1, x2 = s2, . . . , xn = sn is called a solution of linear equationa1x1 + a2x2 + � � �+ anxn = b if a1s1 + a2s2 + � � �+ ansn = b issatis�ed.
Example
Consider the linear equation: 2x1 + x2 = 4.
x1 = 1 and x2 = 2 is a solution.
For any real number t, x1 = t and x2 = 4� 2t is a solution.(Here t is called a parameter.)
Linear Algebra
Systems of Linear Equations
Linear Systems
Exercise
Problem
Assume u = (u1, u2, ..., un) and v = (v1, v2, ..., vn) are twosolutions of a1x1 + a2x2 + ...+ anxn = b. Prove that for any realnumber c, u + c(u � v) is a solution ofa1x1 + a2x2 + ...+ anxn = b.
Solution (Hints)
Show that (u � v) is a solution ofa1x1 + a2x2 + ...+ anxn = 0.
Then, you can prove u + c(u � v) is a solution ofa1x1 + a2x2 + ...+ anxn = b.
Linear Algebra
Systems of Linear Equations
Linear Systems
Linear Systems
De�nition (Linear Systems)
A linear system of m equations in n unknows is a collection of mlinear equations in n common unknowns.8>>><>>>:
a11x1 + a12x2 +...+ a1nxn = b1a21x1 + a22x2 +...+ a2nxn = b2
...am1x1 + am2x2 +...+ amnxn = bm
.
It is called an m� n system. A solution to an m� n system is anordered n-tuple of real numbers (x1, x2, � � � , xn) 1 that satis�es allm equations of the system. The set of all solutions to a linearsystem is called the solution set of the system.
1Here an (ordered) n-tuple of real numbers (x1, x2, � � � , xn) is the same as avector in Rn space.
Linear Algebra
Systems of Linear Equations
Linear Systems
Example
(1, 0) is a solution of�x1 + 3x2 = 4x1 + x2 = 3
. f(1, 0)g is the solutionsset of the linear system.
Example
(0, 0, 3), (1, 0, 2), and (4, 0,�1) are solutions of�x1 +x2 +x3 = 3x1 +x3 = 3
. Actually, this system has in�nite
solution, and its solution set is f(3� t, 0, t) j t 2 Rg.
Example
The system
8<:x1 +x2 = 32x1 +x2 = 33x1 +x2 = 4
has no solution, so its solution set
is ?.
Linear Algebra
Systems of Linear Equations
Linear Systems
The Number of Solutions of A Linear System
1 2 3
3
2
1
1
−=−
=+
13
yxyx
=+
=+
6223
yxyx
1 2 3
3
2
1
1
=+
=+
13
yxyx
1 2 3
3
2
1
1
Linear Algebra
Systems of Linear Equations
Linear Systems
Problem
Prove the number of solutions of a linear system must be one ofthe following cases
1 Exactly 1 solution
2 In�nite number of solutions
3 No solution
(Hint: If u = (u1, u2, � � � , un) and v = (v1, v2, � � � , vn) aresolutions of a linear system in n variables, then for any c 2 R,u+ c (u� v) is also a solution of the system.)
De�nition (Consistent and Inconsistent)
A linear system is inconsistent if its solution set is empty, otherwiseit is consistent.2
2Note: A consistent linear system has either exactly one solution orotherwise in�nite solutions.
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Gaussian Elimination
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Row-Echelon Systems
De�nition (Row-Echelon Form)
A system is in row-echelon form if it follows a stair-step patternand has leading coe¢ cients of 1.
1 All variables are aligned.
2 In an equation, the leading coe¢ cient is the coe¢ cient of the�rst variables.
Example
1 7552
43
932
=+−
−=+−
=+−
zyx
yx
zyx
rowechelon formnot rowechelon form
2
53
932
=
=+
=+−
z
zy
zyx
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Solve a Row-Echelon System
A row-echelon system can be solved by back-substitution.
x � 2y + 3z = 9y + 3z = 5
z = 2row-echelon system
z = 2y = 5� 3z = 5� 3� 2 = �1x = 9+ 2y � 3z = 9+ 2� (�1)� 3� 2 = 1
back-substitution
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Gaussian Elimination
De�nition (Equivalent Systems)
Two systems of linear equations are called equivalent if they haveprecisely the same solution set.
Operations producing equivalent systems
1 Interchange two equations.2 Multiply an equation by a nonzero constant.3 Add a multiple of an equation to another equation.
Gaussian elimination: rewrite a system to an equivalentrow-echelon system by a sequence of these three operations.
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Solve a System of Linear Equations
17552
43
932
=+−
−=+−
=+−
zyx
yx
zyx
17552
53
932
=+−
=+
=+−
zyx
zy
zyx
1
53
932
−=−−
=+
=+−
zy
zy
zyx
42
53
932
=
=+
=+−
z
zy
zyx
2
53
932
=
=+
=+−
z
zy
zyx After applying backsubstitution , we havex = 1, y = 1, z = 2.
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
An mxn Matrix
mnmm
n
n
aaa
aaa
aaa
L
MOMM
L
L
21
22221
11211
m rows
n columns
1st row
2nd row
mst row
1st column nst column
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Represent a Linear System by a Matrix
1752
43
932
=+
−=+−
=+−
zx
yx
zyx
−
−
502031321
−−
−
175024031
9321
system
coefficient matrix
augmented matrix
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Elementary Row Operations
Interchange two rows.
Multiply a row by a nonzero constant.
Add a multiple of a row to another row.
Remark: Note the elementary row operation performed in eachstep.
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Example
21
Notation
RR ↔
−−
−
17552
4031
9321
−
−−
17552
9321
4031
( ) 3312
Notation
RRR ↔+−
−−
−
17552
4031
9321
−−
−−
−
1190
4031
9321
( ) 112
Notation
RR →−
−−
−−−
17552
4031
18642
−−
−
17552
4031
9321
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Row-Echelon Form
Row-echelon form
1 All rows consisting entirely of zeros occur at the bottom of thematrix.
2 For each row that does not consist entirely of zeros, the �rstnonzero entry is 1 (called a leading 1).
3 For two successive (nonzero) rows, the leading 1 in the higherrow is farther to the left than the leading 1 in the lower row.
Reduced row-echelon form
Every column that has a leading 1 has zeros in every positionexcepting the leading 1.
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Gaussian Elimination with Back-Substitution
x �2y +3z = 9�x +3y = �42x �5y +5z = 17
24 1 �2 3 9�1 3 0 �42 �5 5 17
35x �2y +3z = 9
y +3z = 5�y �z = �1
24 1 �2 3 90 1 3 50 �1 �1 �1
35 R2 + R1 ! R2R3 � 2R1 ! R3
x �2y +3z = 9y +3z = 5
2z = 4
24 1 �2 3 90 1 3 50 0 2 4
35 R3 + R2 ! R3
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Gaussian Elimination with Back-Substitution
x �2y +3z = 9y +3z = 5
z = 2
24 1 �2 3 90 1 3 50 0 1 2
35 12R3 ! R3
z = 2y = 5� 3z = 5� 3x2 = �1x = 9+ 2y � 3z = 9+ 2x(�1)� 3x2 = 1
9=;Back-Substitution
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Practice
y + z � 2w = �3x + 2y � z = 22x + 4y + z � 3w = �2x � 4y � 7z � w = �19
26640 1 1 �2 �31 2 �1 0 22 4 1 �3 �21 �4 �7 �1 �19
3775Hint: Exchange the 1st and 2nd equations.
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
Solution26640 1 1 �2 �31 2 �1 0 22 4 1 �3 �21 �4 �1 �7 �19
3775 !26641 2 �1 0 20 1 1 �2 �32 4 1 �3 �21 �4 �1 �7 �19
3775 !26641 2 �1 0 20 1 1 �2 �30 0 3 �3 �60 �6 �6 �1 �21
3775 !26641 2 �1 0 20 1 1 �2 �30 0 3 �3 �60 0 0 �13 �39
3775 !26641 2 �1 0 20 1 1 �2 �30 0 1 �1 �20 0 0 �13 �39
3775 !26641 2 �1 0 20 1 1 �2 �30 0 1 �1 �20 0 0 1 3
3775
Linear Algebra
Systems of Linear Equations
Gaussian Elimination
The Number of Solutions
Inconsistent system: a row with zeros except for the last entry(example)
Consistent system: not inconsistent systems
1 One solution: the number of not-zero rows is equal to thenumber of variables
2 In�nite solutions: the number of not-zero rows is less than thenumber of variables
Quiz: Homogeneous systems in which each of the constantterm is zero are consistent.
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
Gauss-Jordan Elimination
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
Gauss-Jordan Elimination
Continues the procedure of Gaussian elimination until areduced row-echelon form is obtained. For example,
x �2y +3z = 9�x +3y = �42x �5y +5z = 1724 1 �2 3 9
�1 3 0 �42 �5 5 17
35!24 1 �2 3 90 1 3 50 �1 �1 �1
35!24 1 �2 3 90 1 3 50 0 2 4
35!24 1 �2 3 90 1 3 50 0 1 2
35!24 1 �2 0 30 1 0 �10 0 1 2
35!24 1 0 0 10 1 0 �10 0 1 2
35
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
Example
y + z � 2w = �3x + 2y � z = 22x + 4y + z � 3w = �2x � 4y � 7z � w = �19
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
Solution26640 1 1 �2 �31 2 �1 0 22 4 1 �3 �21 �4 �7 �1 �19
3775! � � � !
26641 2 �1 0 20 1 1 �2 �30 0 1 �1 �20 0 0 1 3
3775
!
26641 2 �1 0 20 1 1 0 30 0 1 0 10 0 0 1 3
3775!26641 2 0 0 30 1 0 0 20 0 1 0 10 0 0 1 3
3775
!
26641 0 0 0 �10 1 0 0 20 0 1 0 10 0 0 1 3
3775
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
More Examples
72332
2342
22
32
=+−+
−=−++
=−+
−=−+
wzyx
wzyx
zyx
wzy
twx
twy
twz
tw
−=−=
+−=+−=
+−=+−=
=
−−
−−→
−−
−−
−
→
−−
−−
−
→
−−
−−
−
→
−−
−−
−−
−
→
−
−−
−−
−
→
−
−−
−
−−
22
11
22
Then,.Let
00000
21100
11010
21001
00000
21100
11010
01021
00000
21100
32110
20121
00000
63300
32110
20121
32110
63300
32110
20121
72332
23142
32110
20121
72332
23142
20121
32110
+−
+−
−
=
t
t
t
t
w
z
y
x
2
1
2
:Ans
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
Example
x + 2y � z = 22x + 4y + z � 6w = �2
Solution�1 2 �1 0 22 4 1 �6 �2
�! ...!
�1 2 0 �2 00 0 1 �2 �2
�Let y = s and w = t, then,
z = �2+ 2w = �2+ 2tx = �2y + 2w = �2s + 2t
Ans:
2664xyzw
3775 =2664�2s + 2t
s�2+ 2t
t
3775
Linear Algebra
Systems of Linear Equations
Gauss-Jordan Elimination
Algorithm for Gaussian Elimination
Input an mxn matrix Ai = 1; j = 1;while i < m and j < nif all entries A(i , j),A(i + 1, j), � � � ,A(m, j) are zerothen j = j + 1 and continue
elseif A(i , j) = 0then 9k > i s.t. A(k, j) 6= 0 and switch row(i) and row(k)
//Here we have A(i , j) 6= 0divide row(i) by A(i , j)for k = i + 1 to mif A(k, j) 6= 0then minus A(k, j) times of row(i) from row(k)
i = i + 1; j = j + 1;
Linear Algebra
Systems of Linear Equations
Applications of Linear Systems
Applications of Linear Systems
Linear Algebra
Systems of Linear Equations
Applications of Linear Systems
Polynomial Curve Fitting
Given n points:(x1, y1), (x2, y2), � � � , (xn, yn).Find a polynomial equationp(x) =a0 + a1x + ...+ an�1xn�1
passing through all n points.
FFF n points give nequations and therefore cansolve n variables,a0, a1, � � � , an�1.
Insert Figure.
Linear Algebra
Systems of Linear Equations
Applications of Linear Systems
Example
Find the polynomial p(x) = a0 + a1x + a2x2 that passes throughthe points (1, 4), (2, 0), and (3, 12).
Solution
From (1, 4), we have a0 + a1 + a2 = 4.From (1, 4), we have a0 + a1 + a2 = 4.From (3, 12), we have a0 + 3a1 + 9a2 = 12.24 1 1 1 41 2 4 01 3 9 12
35 ! ...!
24 1 0 0 240 1 0 �280 0 1 8
35Answer: p(x) = 24� 28x + 8x2.
Linear Algebra
Systems of Linear Equations
Applications of Linear Systems
Exercise
1 Find the polynomial p(x) = a0 + a1x + a2x2 + a3x3 + a4x4
passing through the points (�2, 3), (�1, 5), (0, 1), (1, 4),(2, 10). 266664
1 �2 (�2)2 (�2)3 (�2)4 31 �1 (�1)2 (�1)3 (�1)4 51 0 02 03 04 11 1 12 13 14 41 2 22 23 24 10
3777752 Find a polynomial p(x) = a0 + a1x + a2x2 + a3x3 passingthrough the points (0, 1), (1, 2), (�1, 0), (2, 3).
Linear Algebra
Systems of Linear Equations
Applications of Linear Systems
Unit Review
Linear equations and systems of linear equations
1 Solutions2 Number of solutions
Gaussian elimination
1 Equivalent systems
elementary row operation
2 Row-echelon form3 Back-substitution
Gauss-Jordan elimination
Reduced row-echelon form
Parametric solutions
Matrix representation
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