Chapter 1
Systems of Linear Equations
1.1 Introduction to Systems of Linear Equations
1.2 Gaussian Elimination and Gauss-Jordan Elimination
1.3 Applications of Systems of Linear Equations
※ The linear algebra arises from solving systems of linear equations
※ The last section of each chapter illustrates how to apply the
knowledge you learn in that chapter to solving problems in
different fields
1.1
1.1 Introduction to Systems of Linear Equations

A linear equation (線性方程式) in n variables:
a1 x1  a2 x2  a3 x3 
 an xn  b
ai : real-number coefficients
xi : variables needed to be solved
b : real-number constant term
a1: leading coefficient (領先係數)
x1: leading variable (領先變數)

Notes:
(1) Linear equations have no products or roots of variables and
no variables involved in trigonometric, exponential, or
logarithmic functions
(2) Variables appear only to the first power
1.2

Ex 1: Linear or Nonlinear
1
x  y  z  2
2
Linear (a) 3 x  2 y  7
(b)
Linear (c) x1  2 x2  10 x3  x4  0
(d) (sin ) x1  4 x2  e 2 Linear
Linear
x is the exponent
Nonlinear (e) xy  z  2
(f) e x  2 y  4
Nonlinear
1 1
(h)   4
x y
Nonlinear
product of variables
Nonlinear (g) sin x1  2 x2  3x3  0
trigonometric function
not the first power
1.3

A solution (解) of a linear equation in n variables:
a1 x1  a2 x2  a3 x3 
 an xn  b
x1  s1 , x2  s2 , x3  s3 , , xn  sn
s.t.

a1s1  a2 s2  a3 s3    an sn  b
Solution set (解集合):
the set of all solutions of a linear equation
※ In most cases, the number of solutions of a linear equation is
infinite (無限大), so we need some methods to represent the
solution set
1.4

Ex 2: Parametric representation (參數化表示) of a solution set
x1  2 x2  4 with a solution (2, 1), i.e., x1  2, x2  1
※ If you solve for x1 in terms of x2, you obtain
x1  4  2 x2 (in this form, the variable x2 is free)
By letting x2  t (the variable t is called a parameter), you can
represent the solution set as
x1  4  2t , x2  t , t is any real number
The set representation for solutions: (4  2t, t ) | t  R
※ If you choose x1 to be the free variable, the parametric
representation of the solution set is ( s, 2  12 s) | s  R
1.5

A system of m linear equations in n variables:
a11 x1 
a21 x1 
a31 x1 
a12 x2
a22 x2
a32 x2
am1 x1  am 2 x2




a13 x3
a23 x3
a33 x3






 am3 x3

 amn xn
a1n xn
a2 n xn
a3n xn



b1
b2
b3
 bm
A solution of a system of linear equations (線性方程式系
統) is a sequence of numbers s1, s2,…,sn that can solve each
linear equation in the system
1.6

Notes:
Every system of linear equations has either
(1) exactly one solution
(2) infinitely many solutions
(3) no solution


Consistent (一致性(有解)):
A system of linear equations has at least one solution (for
cases (1) and (2))
Inconsistent (非一致性(無解、矛盾)):
A system of linear equations has no solution (for case (3))
1.7

Ex 4: Solution of a system of linear equations in 2 variables
(1)
(2)
(3)
x  y  3
x  y  1
two intersecting lines
exactly one solution
x  y  3
2x  2 y  6
two coincident lines
inifinitely many solution
x  y  3
x  y  1
two parallel lines
no solution
1.8

Ex 5: Using back substitution (倒序替換法) to solve a
system in row-echelon form (列梯形形式)
x  2y 
5
y  2
(1)
(2)
※ The row-echelon form means that it follows a stair-step pattern (two
variables are in Eq. (1) and the first variable x is the leading variable, and one
variable is in Eq. (2) and the second variable y is the leading variable) and
has leading coefficients of 1 for all equations
※ The back substitution means to solve a system in reverse order. That is, you
solve Eq. (2) for y first, and then substitute the solution of y into Eq. (1).
Next, you can solve x directly in Eq. (1) since x is the only unknown variable
Sol: By substituting y  2 into Eq. (1), you obtain
x  2(2)  5  x  1
The system has exactly one solution: x  1, y  2
1.9

Ex 6: Using back substitution to solve a system in rowechelon form
x  2 y  3z  9
(1)
y  3z  5
(2)
z  2
(3)
Sol: Substitute z  2 into (2), y can be solved as follows
y  3(2)  5
y  1
and substitute y  1 and z  2 into (1)
x  2( 1)  3(2)  9
x  1
The system has exactly one solution:
x  1, y  1, z  2
1.10


Equivalent (等價):
Two systems of linear equations are called equivalent
if they have precisely the same solution set
Notes:
Each of the following operations (運算) on a system of linear
equations produces an equivalent system
O1: Interchange two equations
O2: Multiply an equation by a nonzero constant
O3: Add a multiple of an equation to another equation

Gaussian elimination:
A procedure to rewrite a system of linear equations to be in
row-echelon form by using the above three operations
1.11

Ex 7: Solve a system of linear equations (consistent system)
x  2 y  3z  9
(1)
x  3y
 4
(2)
2 x  5 y  5 z  17
(3)
Sol: First, eliminate the x-terms in Eqs. (2) and (3) based
on Eq. (1)
(1)  (2)  (2) (by O3)
x  2 y  3z  9
y  3z  5
2 x  5 y  5 z  17
(1)  (2)  (3)  (3) (by O3)
x  2 y  3z  9
y  3z  5
 y  z  1
(4)
(5)
1.12
Second, eliminate the (-y)-term in Eq. (5) based on Eq. (4)
(4)  (5)  (5) (by O3)
x  2 y  3z  9
y  3z  5
2z  4
(6)
(6)  12  (6) (by O2)
x  2 y  3z  9
y  3z  5
z  2
Since the system of linear equations is expressed in its row-
echelon form, the solution can be derived by the back
substitution: x  1, y  1, z  2 (only one solution)
1.13

Ex 8: Solve a system of linear equations (inconsistent system)
x1  3x2  x3  1
2 x1  x2  2 x3  2
x1  2 x2  3x3   1
(1)
(2)
(3)
Sol:
(1)  (  2)  (2)  (2) (by O3)
(1)  (  1)  (3)  (3) (by O3)
x1  3x2 
x3 
1
5 x2
5 x2
 4 x3
 4 x3
 0
 2
(4)
(5)
1.14
(4)  (1)  (5)  (5) (by O3)
x1  3x2 
x3 
1
5 x2  4 x3  0
0  2
(a false statement)
So, the system has no solution (an inconsistent system)
1.15

Ex 9: Solve a system of linear equations (infinitely many solutions)
x2
x1
 x1
Sol:
 x3
 3 x3
 3 x2
(1)  (2) (by O1)
x1
 3 x3
x2  x3
 x1  3 x2
 0
 1
 1
(1)
(2)
(3)
 1
 0
 1
(1)
(2)
(3)
(1)  (3)  (3) (by O3)
x1
 3 x3  1
x2  x3  0
3 x2  3 x3  0
(4)
1.16
Since Equation (4) is the same as Equation (2), it is not
necessary and can be omitted
x1
x2
 3x3
 x3


1
0
 x2  x3 , x1  1  3x3
Let x3  t , then x1  3t  1
x2  t
tR
x3  t
This system has infinitely many solutions.
1.17
Keywords in Section 1.1:

linear equation: 線性方程式

system of linear equations: 線性方程式系統

leading coefficient: 領先係數

leading variable: 領先變數

solution: 解

solution set: 解集合

free variable: 自由變數

parametric representation: 參數化表示

consistent: 一致性 (有解)

inconsistent: 非一致性 (無解、矛盾)

row-echelon form: 列梯形形式

equivalent: 等價
1.18
1.2 Gaussian Elimination and Gauss-Jordan Elimination

mn matrix (矩陣):
 a11
 a21
 a31


 a m1
a12
a22
a32

am 2
a13  a1n 
a23  a2 n 
a33  a3n 


am 3  amn 
m rows (列)
n columns (行)
Notes:
(1) Every entry (元素) aij in the matrix is a real number
(2) A matrix with m rows and n columns is said to be with size mn

(3) If m  n, then the matrix is called square matrix (方陣) of order n
(4) For a square matrix, the entries a11, a22, …, ann are called the
main (or principal) diagonal (主對角線) entries
1.19


Ex 1:
Matrix
[ 2]
Size
1 1 ※ A scalar (any real number)
can be treated as a 11 matrix.
 0 0
0 0
22
1  3 0 1 

2 
1 4
 e  
2
2
  7 4 
3 2
Note:
One very common use of matrices is to represent a system
of linear equations (see the next slide)
1.20

A system of m equations in n variables:
a11 x1
a21 x1
a31 x1



am1 x1
 am 2 x2
a12 x2
a22 x2
a32 x2



a13 x3
a23 x3
a33 x3






 am3 x3

 amn xn
 bm
a13  a1n 
a23  a2 n 
a33  a3n 


am 3  amn 
 x1 
x 
x   2
 
 
 xn 
 b1 
b 
b 2
 
 
bm 
a1n xn
a2 n xn
a3n xn



b1
b2
b3
Matrix form: Ax  b
 a11
 a21
A   a31


 a m1
a12
a22
a32

am 2
1.21

Coefficient matrix (係數矩陣):
 a11 a12
a
 21 a22
 a31 a32


 am1 am 2

a13
a23
a33
am3
a1n 
a2 n 
a3n   A


amn 
Augmented matrix (增廣矩陣): formed by appending the
constant-term vector to the right of the coefficient matrix of a
system of linear equations
 a11 a12
a
 21 a22
 a31 a32


 am1 am 2
a13
a23
a33
a1n
a2 n
a3n
am3
amn
b1 
b2 
b3   [ A b]


bm 
1.22


Before introducing the Gaussian elimination and the GaussJordan elimination, we need some background knowledge,
including the elementary row operations (基本列運算) and
the rules to identify the row-echelon or reduced row-echelon
forms (簡化列梯形形式) for matrices
Elementary row operations (基本列運算):
(1) Interchange two rows: I i , j  Ri  R j
(2) Multiply a row by a nonzero constant: M i( k )  (k ) Ri  Ri
(3) Add a multiple of a row to another row: Ai(,kj)  (k ) Ri  R j  R j

Row equivalent (列等價):
Two matrices are said to be row equivalent if one can be obtained
from the other by a finite sequence of above elementary row
operations
1.23

Ex 2: Elementary row operations
 0 1 3 4
 1 2 0 3
 2  3 4 1
 2  4 6  2
 1 3  3 0
 5  2 1 2
 1 2  4 3
0 3  2  1
2 1 5  2
I1,2
M
1
( )
2
1
( 2)
A1,3
 1 2 0 3
 0 1 3 4
 2  3 4 1
 1  2 3  1
 1 3  3 0
5  2 1 2
 1 2  4 3
0 3  2  1
0  3 13  8
1.24

Row-echelon form (列梯形形式): (1), (2), and (3)

Reduced row-echelon form (簡化列梯形形式): (1), (2), (3), and (4)
(1) All rows consisting entirely of zeros occur at the bottom of the
matrix (若有全部為0之row，會出現在matrix的最下方)
(2) For each row that does not consist entirely of zeros, the first
nonzero entry from the left side is 1, which is called as leading 1
(對不全為0之row，從左到右，第一個出現非0的entry，其
值為1，且此entry稱為leading 1)
(3) For two successive nonzero rows, the leading 1 in the higher
row is further to the left than the leading 1 in the lower row (對
兩相鄰不全為0之row，越上方的row中的leading 1，會在越
左邊)
(4) Every column that contains a leading 1 has zeros everywhere
else (每個有leading 1的column中，其它的entry均為0)
1.25

Ex 4: Row-echelon form or reduced row-echelon form
 1 2  1 4
0 1 0 3
0 0 1  2
(row-echelon
form)
0 1 0 5
0 0 1 3
0 0 0 0
0  1
0 2
1 3
0 0
 1  5 2  1 3
0 0 1 3  2 (row-echelon
0 0 0 1 4 form)
0 0 0 0 1
1
0
0
0
 1 2  3 4
0 2 1  1
0 0 1  3
 1 2  1 2
 0 0 0 0
0 1 2  4
Violate the
second criterion
0
1
0
0
(reduced rowechelon form)
(reduced rowechelon form)
Violate the first
criterion
1.26



Gaussian elimination:
The procedure for reducing a matrix to a row-echelon form
by performing the three elementary row operations
Gauss-Jordan elimination:
The procedure for reducing a matrix to its reduced row-echelon
form by performing the three elementary row operations
Notes:
(1) Every matrix has an unique reduced row-echelon form
(2) A row-echelon form of a given matrix is not unique
(Different sequences of elementary row operations can
produce different row-echelon forms)
※ The above two statements will be verified numerically in Ex. 8
1.27

Ex: Procedure of the Gaussian elimination and Gauss-Jordan elimination
Produce leading 1
0 0  2 0 7 12
2 4  10 6 12 28
2 4  5 6  5  1
I1,2
2 4  10 6 12 28
0 0  2 0 7 12
2 4  5 6  5  1
The first nonzero column
Leading 1
M 1  1 2  5 3 6 14
0 0  2 0 7 12
2 4  5 6  5  1
( 12 )
Produce leading 1
( 2)
6 14
1 2  5 3
A1,3
7 12
0 0  2 0
0 0 5 0  17  29
Nonzero entries below leading 1
The first nonzero
column
Submatrix
1.28
The first nonzero column
Leading 1
M
(  12 )
2
6 14 
 1 2 5 3
0 0

1
0

7
2

6


0 0 5 0 17 29 
6 14 
 1 2 5 3
0 0

1
0

7
2

6


0 0 0 0
12
1
( 5)
2,3
A
Nonzero entries below leading 1
Produce Submatrix
leading 1
Non-zeros
M
(2)
3
6 14 
 1 2 5 3
0 0

1
0

7
2

6


0 0 0 0
1 2 
( 6)
3,1
A
( 72 )
3,2
A
(5)
A2,1
 1 2 0 3 0 7
0 0 1 0 0 1
0 0 0 0 1 2
Leading 1
(row-echelon form)
(reduced rowechelon form)
1.29

Ex 7: Solve a system by the Gauss-Jordan elimination method (only
one solution)
x  2 y  3z  9
 x  3y
 4
2 x  5 y  5z  17
Sol:
Form the augmented matrix by concatenating A and b
 1 2 3 9  A(1) A( 2)  1 2 3 9  A(1) M (1/ 2)  1 2 3 9 
2,3
3
 1 3 0 4 1,2 1,3 0

0

1
3
5
1
3
5






 2 5 5 17 
0 1 1 1
0 0 1 2 
(row-echelon form)
1
( 3)
( 3)
(2)  1 0 0
A3,2 A3,1 A2,1 
x
 1

y
 1
0 1 0 1
0 0 1 2 
z  2
(reduced row-echelon form)
1.30

Ex 8：Solve a system by the Gauss-Jordan elimination
method (infinitely many solutions)
2 x1  4 x2  2 x3  0
3 x1  5 x2
 1
Sol:
For the augmented matrix,
 2 4 2
3 5 0

1
M 2( 1)


0
0  M1( 12 )  1 2 1 0  A1,2( 3)  1 2 1 0 







1
3
5
0
1
0

1
3
1




2 1 0  A2,1( 2)  1 0 5 2 




1 3 1
0
1

3

1


(row-echelon form)
(reduced row-echelon form)
1.31
Then the system of linear equations becomes
x1
 5 x3  2
x2  3x3  1
It can be further reduced to
x1  2  5 x3
x 2   1  3 x3
Let x3  t , then x1  2  5t ,
x2  1  3t ,
tR
x3  t ,
So, this system has infinitely many solutions
1.32
※ In order to show that the reduced row-echelon form is unique, I
conduct an experiment by performing a redundant elementary row
operation of interchanging the first and second rows in advance
 2 4 2
3 5 0

( 2 )
1
A1,2


0
0  I1,2  3 5 0 1 M1(1/ 3)  1 5 / 3 0 1/ 3
 
 



1
2
4

2
0
2
4

2
0




5/3 0
1/ 3 M 2( 3 / 2 )  1 5 / 3 0 1/ 3
 


2 / 3 2 2 / 3
0
1

3

1


 1 0 5 2
 

0
1

3

1


( 5 / 3)
A2,1
(row-echelon form)
(reduced row-echelon form)
※ Comparing with the results on Slide 1.31, we can infer that it is
possible to derive different row-echelon forms, but there is a unique
reduced row-echelon form for each matrix
1.33

Homogeneous systems of linear equations (齊次線性系統) :
A system of linear equations is said to be homogeneous
if all the constant terms are zero
a11 x1  a12 x2 
a21 x1  a22 x2 
a31 x1  a32 x2 

am1 x1  am 2 x2 
a13 x3    a1n xn  0
a23 x3    a2 n xn  0
a33 x3    a3n xn  0
am 3 x3    amn xn  0
1.34

Trivial (obvious) solution (顯然解):
x1  x2  x3    xn  0


Nontrivial solution (非顯然解):
other solutions
Theorem 1.1
(1) Every homogeneous system of linear equations is consistent.
Furthermore, for a homogeneous system, exactly one of the
following is true:
(a) The system has only the trivial solution
(b) The system has infinitely many nontrivial solutions in addition to the
trivial solution (In other words, if a system has any nontrivial solution, this
system must have infinitely many nontrivial solutions)
(2) If the homogenous system has fewer equations than variables,
then it must have an infinite number of solutions
1.35

Ex 9: Solve the following homogeneous system (also verify
the two statements in Theorem 1.1 numerically)
x1  x2
2 x1  x2
 3x3
 3x3
 0
 0
Sol: For the augmented matrix,
( 13 )
( 2)
(1)
A
M
A
1,2
2
2,1
 1  1 3 0
2 1 3 0

x1
x2
 2 x3

x3
 1 0 2 0
0 1  1 0
(reduced rowechelon form)
 0
 0
Let x3  t, then x1  2t , x2  t , x3  t , t  R
When t  0, x1  x2  x3  0 (trivial solution)
When t  0, there are infinitely many nontrivial solutions
1.36
Keywords in Section 1.2:

matrix: 矩陣

row: 列

column: 行

entry: 元素

size: 大小

square matrix: 方陣

order: 階

main diagonal: 主對角線

coefficient matrix: 係數矩陣

augmented matrix: 增廣矩陣
1.37

elementary row operation: 基本列運算

row equivalent: 列等價

row-echelon form: 列梯形形式

reduced row-echelon form: 簡化列梯形形式

leading 1: 領先1

Gaussian elimination: 高斯消去法

Gauss-Jordan elimination: 高斯-喬登消去法

homogeneous system: 齊次系統

trivial solution: 顯然解

nontrivial solution: 非顯然解
1.38
1.3 Applications of Systems of Linear Equations

Polynomial Curve Fitting in Examples 1, 2, and 4
See the text book or “Applications in Ch1.pdf” downloaded
from my website
The polynomial curve fitting problem is that given n points
(x1, y1), (x2, y2),…, (xn, yn) on the xy-plane, find a polynomial
function of degree n–1 passing through these n points
Once equipped with this polynomial function, we can predict
the value of y for some missing values of x
A typical application of the curve fitting problem in the field
of finance is to derive the interest rate (y) for different time to
maturity (x)




1.39