Linear Algebra
Chapter 1
Linear Equations in Linear
Algebra
1.1 Matrices and Systems of
Linear Equations
Definition
• An equation such as x+3y=9 is called a linear equation
(in two variables or unknowns).
• The graph of this equation is a straight line in the xy-plane.
• A pair of values of x and y that satisfy the equation is called
a solution.
Ch1_2
Definition
A linear equation in n variables x1, x2, x3, …, xn has the
form a1 x1 + a2 x2 + a3 x3 + … + an xn = b
where the coefficients a1, a2, a3, …, an and b are real
numbers.
Ch1_3
Solutions for system of linear equations
Figure 1.1
Unique solution
x + 3y = 9
–2x + y = –4
Lines intersect at (3, 2)
Unique solution:
x = 3, y = 2.
Figure 1.2
No solution
–2x + y = 3
–4x + 2y = 2
Lines are parallel.
No point of intersection.
No solutions.
Figure 1.3
Many solution
4x – 2y = 6
6x – 3y = 9
Both equations have the
same graph. Any point on
the graph is a solution.
Many solutions.
Ch1_4
A linear equation in three variables corresponds to
a plane in three-dimensional space.
※ Systems of three linear equations in three variables:
Unique solution
Ch1_5
No solutions
Many solutions
Ch1_6
A solution to a system of a three linear equations will be
points that lie on all three planes.
The following is an example of a system of three linear
equations:
x1 
x2 
x3  2
2 x1  3 x2 
x3  3
x1  x2  2 x3  6
How to solve a system of linear equations? For this we
introduce a method called Gauss-Jordan elimination.
(Section1.2)
Ch1_7
Definition
• A matrix is a rectangular array of numbers.
• The numbers in the array are called the elements of the
matrix.
Matrices
 2 3  4
A

7
5

1


 7 1
B   0 5
 8 3
3 5 6 
C  0  2 5 
8 9 12
Ch1_8
 2 3  4
A

7
5

1


Row and Column
2
3  4
7
row 1
5  1
row 2
 2
7 
 
column 1
3
5
 
column 2
  4
  1
 
column 3
Submatrix
1 7 4 
A  2 3 0 
5 1  2
matrix A
 1 7
P  2 3
 5 1
7 
4
1


Q   3
R

5

2


1 
submatrices of A
Ch1_9
Size and Type
 1 0
 2 4

 2 5 7
 9 0 1 


  3 5 8 
3  3 matrix
3
5
Size : 2  3
1 4 matrix
8 
 3
 
2
3  1 matrix
a row matrix
a column matrix
4
a square matrix
3 8
5
Location
 2 3  4
A

7 5  1
a13  4, a21  7
The element aij is in row i , column j
The element in location (1,3) is 4
Identity Matrices
diagonal size
I 2  1 0 
0 1
1 0 0 
I 3  0 1 0 
0 0 1
Ch1_10
Relations between system of linear equations
and matrices
matrix of coefficients and augmented matrix
x1  x2  x3  2
2 x1  3 x2  x3  3
x1  x2  2 x3  6
1
1 1
2

3
1


 1  1  2
matrix of coefficients
1
2
1 1
2

3
1
3


 1  1  2  6
augmented matrix
Ch1_11
Elementary Row Operations of Matrices
Elementary Transformation
1. Interchange two equations.
2. Multiply both sides of an
equation by a nonzero
constant.
3. Add a multiple of one
equation to another equation.
Elementary Row
Operation
1. Interchange two rows of a
matrix.
2.
Multiply the elements of a
row by a nonzero constant.
3.
Add a multiple of the
elements of one row to the
corresponding elements of
another row.
Ch1_12
Example 1
Solving the following system of linear equation.
x1  x2  x3  2
2 x1  3 x2  x3  3
x1  x2  2 x3  6
 row equivalent
Solution
Equation Method
Initial system:
x1  x2  x3  2
Eq2+(–2)Eq1
2 x1  3 x2  x3  3
x1  x2  2 x3  6
Eq3+(–1)Eq1
x1  x2  x3  2
Analogous Matrix Method
Augmented matrix:
1
2
1 1
3
1
3
2
 1  1  2  6

x2  x3  1 R2+(–2)R1
 2 x2  3 x3  8 R3+(–1)R1
1
1
2
1
1  1  1
0
0  2  3  8
Ch1_13
Eq1+(–1)Eq2
Eq3+(2)Eq2
(–1/5)Eq3
Eq1+(–2)Eq3
Eq2+Eq3
x1  x2  x3  2
x2  x3  1
 2 x2  3 x3  8
1
1
2
1
1  1  1
0
0  2  3  8
x1  2 x3  3
x2  x3  1
 5 x3  10

R1+(–1)R2
R3+(2)R2
2
3
1 0
0 1  1  1
0 0  5  10
x1  2 x3  3
x2  x3  1
x3  2

(–1/5)R3
2
3
1 0
0 1  1  1
1
2
0 0
The solution is
x1  1, x2  1, x3  2.
x1  1
x2  1
x3  2

R1+(–2)R3
R2+R3
1
0


0
0
0
1
0
0
1
The solution is
x1  1, x2  1, x3  2.
 1
1

2

Ch1_14
Example 2
Solving the following system of linear equation.
x1  2 x2  4 x3  12
2 x1  x2  5 x3  18
 x1  3 x2  3 x3  8
Solution

4 12
4 12
 1 2
1  2
R2  (2)R1
5 18
3  3  6
 2 1
0
R3  R1
3  3  8
1
1
4
 1
0

1
 R2
3
 1  2 4 12
1  1  2
0
1 1
4
0
8
1 0 2

0 1  1  2 R1  (2)R3
1

 R2  R3
 R3
0 0
1
3
2


R1  (2)R2
R3  (1)R2
8
1 0 2
0 1  1  2 
6
0 0 2
 1 0 0 2
0 1 0 1


0 0 1 3
 x1  2

solution  x2  1.
x  3
 3
Ch1_15
Example 3
4 x1  8 x2  12 x3  44
Solve the system
3x1  6 x2  8 x3  32
 2 x1  x2
 7
Solution
 4 8  12 44

 3 6  8 32

  1 R1
 2  1
0  7   4 
 1 2  3 11
 1 2  3 11

 3 6  8 32 R2  (3)R1 

0
0
1

1




R3

2
R1
 2  1
0  7 
0 3  6 15

 1 2  3 11


R2  R3 0 3  6 15  1 
 R2
0 0
1  1  3 


1 0 0
R1  (1)R3
R2  2R3
2
0 1 0
.
3


0 0 1  1
 1 2  3 11

0 1  2

5

 R1  (2)R2
0 0
1  1
1 1
1 0
0 1  2 5


0 0
1  1
The solution is x1  2, x2  3, x3  1.
Ch1_16
Summary
8  12 44
 4
 3 6  8 32
[
A
:
B
]

3x1  6 x2  8 x3  32


 2  1
0  7 
 2 x1  x2
 7
A
B
Use row operations to [A: B] :
4 x1  8 x2  12 x3  44
 4 8  12 44
 3 6  8 32   


 2  1
0  7 
1
0

0
0
0
1
0
0
1
2
3. i.e., [ A : B ]    [ I n
 1
: X]
Def. [In : X] is called the reduced echelon form of [A : B].
Note. 1. If A is the matrix of coefficients of a system of n equations
in n variables that has a unique solution,
then A is row equivalent to In (A  In).
2. If A  In, then the system has unique solution.
Ch1_17
Example 4 Many Systems
Solving the following three systems of linear equation, all of
which have the same matrix of coefficients.
x1  x2  3x3  b1
 b1   8 0  3
2 x1  x2  4 x3  b2 for b2    11,  1,  3 in turn
b   11 2  4
 x1  2 x2  4 x3  b3
 3 
    
Solution

3
8 0
3
3
8 0
3
 1 1
 1 1
4
11 1
3 R2+(–2)R1 0
1  2  5 1  3
 2 1
R3+R1
1  1  3 2  1
 1 2  4  11 2  4
0
1 0
1
3 1
0
  1 0 0 1 0  2



0 1  2  5 1  3 R1 ( 1) R3 
R1 R2
0 1 0 1 3
1




R3 ( 1)R2
0 0
 R2  2 R3 
1
2
1
2
2 1
2


0 0 1
 x1  1  x1  0  x1  2
The solutions to



x


1
,
x

3
,
 2
 x2  1 .
the three systems are  2
x  2 x  1 x  2
 3
 3
 3
Ch1_18
Homework
Exercises will be given by the teachers of
the practical classes.
Ch1_19