Module 2 Chapter 12 System of Linear Equations

1
12
System of Linear Equations
Case Study
12.1 System of Linear Equations
12.2 Solving Equations by Inverses of Matrices
12.3 Solving Equations by Cramers Rule
12.4 Solving Equations by Gaussian Elimination
12.5 Homogeneous Systems of Linear Equations
Chapter Summary
2
Case Study
In a Biology lesson, a group of students are
doing experiments to study the process of
photosynthesis.
During the process, carbon dioxide (CO2) and
water (H2O) would be converted into glucose
(C6H12O6), and some oxygen (O2) is released
p CO2 ? q H2O ? r C6H12O6 ? s O2 where p, q,
r and s are real numbers.
In order to balance the equation, the numbers of
atoms of carbon (C), oxygen (O) and hydrogen (H)
should be the same on both sides of the equation.
For example Number of carbon atoms before the
process ? p
? p ? 6r
Number of carbon atoms after the
process ? 6r
3
Case Study
Chemical equation p CO2 ? q H2O ? r C6H12O6
? s O2 where p, q, r and s are real numbers.
Number of C atoms before the process ? p
? p ? 6r
Number of C atoms after the process ? 6r
Number of O atoms before the process ? 2p q
? 2p q ? 6r 2s
Number of O atoms after the process ? 6r 2s
Number of H atoms before the process ? 2q
? 2q ? 12r
Number of H atoms after the process ? 12r
We can express the above details as a system of
linear equations
4
12.1 System of Linear Equations
A system of m linear equations (or a linear
system) in n unknowns x1, x2, x3, ¼, xn is a set
of equations of the form
The constants aij are called the coefficients of
the system of linear equations.
For example, is a
system of two linear equations with three
unknowns x, y and z.
5
12.1 System of Linear Equations
In a system of linear equations, if there exists
a set of numbers N1, N2, , Nn satisfying all
the equations, then the system is said to be
solvable or consistent, and
N1, N2, , Nn is called a solution of the
system of linear equations.
Otherwise, the system is said to be non-solvable
or inconsistent.
The system of linear equations may be represented
by the matrix equation
Here, A is called the coefficient matrix, x is
called the unknown matrix and (N1, N2, , Nn)t is
called the solution matrix.
6
12.2 Solving Equations by Inverses of
Matrices
Suppose we have a system of linear equations of
order 3
We can express the system in the matrix equation
Ax ? b, where A is a 3 ? 3 coefficient matrix.
If A is non-singular, then the solution matrix x
can be found by computing the inverse of A, and
the solution is unique.
7
12.2 Solving Equations by Inverses of
Matrices
Theorem 12.1 Let A be a square matrix. If A is
non-singular, then the system of linear equations
Ax ? b has a unique solution given by x ? A?1b.
Proof If A is a non-singular matrix, then A?1
exists.
Ax ? b
(A?1A)x ? A?1b
Ix ? A?1b
x ? A?1b
Therefore, the solution of Ax ? b exists.
Now suppose Ax ? b has two solutions x1 and x2.
Then Ax1 ? b and Ax2 ? b.
x1 ? A?1b and x2 ? A?1b.
\ x1 ? x2
Therefore, the solution of Ax ? b is unique.
8
12.2 Solving Equations by Inverses of
Matrices
Example 12.1T
Solution
\ The unique solution of the system of linear
equations is x ? 18, y ? 10.
9
12.2 Solving Equations by Inverses of
Matrices
Example 12.2T
Solution
10
12.2 Solving Equations by Inverses of
Matrices
We know that a square matrix A is non-singular if
and only if A ? 0.
Theorem 12.2 Let A be a square matrix. The system
of linear equations Ax ? b has a unique solution
if and only if A ? 0.
This theorem can be used to test whether a system
of linear equations has a unique solution.
When A ? 0, A?1 does not exist, so the method
of inverse matrix cannot be applied.
In this situation, either of the following cases
will happen 1. the system of equations does
not have any solution, or 2. the system of
equations has infinitely many solutions.
11
12.2 Solving Equations by Inverses of
Matrices
Example 12.3T
Solution
(a) Rewrite the system of equations as
\ The system does not have a unique solution.
Since (2) and (3) are the same, we say that
equation (2) is redundant and the linear system
has only one equation x 2y ? 1.
Therefore, the system of linear equation has
infinitely many solutions.
12
12.2 Solving Equations by Inverses of
Matrices
Example 12.3T
Determine the number of solutions to the
following systems of linear equations.
(a) (b)
Solution
(b) Rewrite the system of equations as
\ The system does not have a unique solution.
(3) (2) 0 ? 5, which is impossible.
Therefore, the system of linear equations has no
solution.
13
12.3 Solving Equations by Cramers Rule
Finding the inverse of the coefficient matrix is
sometimes complicated, so in this section we will
study how to use Cramers rule to solve a system
of linear equations in a more convenient way.
14
12.3 Solving Equations by Cramers Rule
If we express x1 and x2 in determinant form, we
can obtain
15
12.3 Solving Equations by Cramers Rule
Example 12.4T
Solution
The determinant of the coefficient matrix
\ The unique solution of the system of linear
equations is
16
12.3 Solving Equations by Cramers Rule
For systems of linear equations of order 3,
Cramers rule is stated as follows
Comparing Theorems 12.3 and 12.4, we can see that
in both cases, the solution xj can be expressed
as a fraction with A as the denominator, and
the numerator is the determinant that replaces
the elements in the jth column of A by bis.
17
12.3 Solving Equations by Cramers Rule
Example 12.5T
Solution
\ The unique solution of the system of linear
equations is
18
12.3 Solving Equations by Cramers Rule
So it is only applicable when the coefficient
matrix A is a non-singular matrix.
19
12.3 Solving Equations by Cramers Rule
Example 12.6T
Suppose we have a system of linear equations
where a, b and c are real numbers. (a) If the
system of linear equations has a unique solution,
show that a, b and c are distinct and a
b c ? 0.
Solution
(a) The determinant of the coefficient matrix
?
20
12.3 Solving Equations by Cramers Rule
Example 12.6T
Suppose we have a system of linear equations
where a, b and c are real numbers. (a) If the
system of linear equations has a unique solution,
show that a, b and c are distinct and a
b c ? 0.
Solution
? The system of linear equations has a unique
solution.
(a)
\ D ? 0
i.e., abc(b a)(c a)(c b)(a b c) ? 0
\ a, b, c are distinct and a b c ? 0.
21
12.3 Solving Equations by Cramers Rule
Example 12.6T
Suppose we have a system of linear equations
where a, b and c are real numbers. (a) If the
system of linear equations has a unique solution,
show that a, b and c are distinct and a
b c ? 0.
(b) Solve the system of linear equations if it
has a unique solution.
Solution
(b)
22
12.3 Solving Equations by Cramers Rule
Example 12.6T
Suppose we have a system of linear equations
where a, b and c are real numbers. (a) If the
system of linear equations has a unique solution,
show that a, b and c are distinct and a
b c ? 0.
(b) Solve the system of linear equations if it
has a unique solution.
Solution
(b)
23
12.3 Solving Equations by Cramers Rule
Example 12.6T
Suppose we have a system of linear equations
where a, b and c are real numbers. (a) If the
system of linear equations has a unique solution,
show that a, b and c are distinct and a
b c ? 0.
(b) Solve the system of linear equations if it
has a unique solution.
Solution
(b)
The unique solution of the system of linear
equations is
24
12.4 Solving Equations by Gaussian
Elimination
In the last two sections, we learnt how to solve
systems of linear equations of order 2 and 3.
However, those methods can only be applied when
the coefficient matrix is a non-singular square
matrix.
If the linear system has an infinite number of
solutions, we cannot find the solutions using
those methods.
Therefore, in this section, we will learn a
general method for solving systems of linear
equations.
Before introducing the method, we first define
the row echelon form for a linear system
25
12.4 Solving Equations by Gaussian
Elimination
The row echelon form of a system of linear
equations has the following characteristics
1. The system contains n unknowns x1, x2, x3, ,
xn.
2. The first non-zero term of each row has a
coefficient of 1.
3. In any two successive rows, for example, the
ith and (i 1)th rows, if the ith row does not
consist entirely of zero terms, then the number
of leading zeros in the (i 1)th row must be
greater than the number of leading zeros in the
ith row.
26
12.4 Solving Equations by Gaussian
Elimination
If a system of equations is given, we can perform
any of the following three elementary
transformations to transform it into the row
echelon form, without affecting the solution of
the system
1. interchanging the position of two
equations, 2. multiplying both sides of an
equation by a non-zero number, 3. adding an
arbitrary multiple of any equation to another
equation.
27
12.4 Solving Equations by Gaussian
Elimination
28
12.4 Solving Equations by Gaussian
Elimination
This process of transforming a system into row
echelon form is called Gaussian elimination.
As shown above, the value of z can be found
directly from the third equation, i.e., z ? 3.
By substituting the value of z into the second
equation, we can find the value of y.
Finally, x can be solved by substituting the
values of y and z into the first equation.
This process is called back-substitution.
29
12.4 Solving Equations by Gaussian
Elimination
Example 12.7T
Solution
(a) Interchange the 1st equation and the 2nd
equation, we have
Add (?1) ? the 3rd equation to the 2nd equation,
we have
Multiply the 1st equation by 1, we have
Add (?5) ? the 1st equation to the 3rd equation,
we have
30
12.4 Solving Equations by Gaussian
Elimination
Example 12.7T
Solution
Multiply the 3rd equation by , we have
(a) Add (?2) ? the 2nd equation to the 3rd
equation, we have
Multiply the 2nd equation by , we have
which is the row echelon form of (E).
31
12.4 Solving Equations by Gaussian
Elimination
Example 12.7T
Solution
(b) From (3), we have z ? 5.
Substituting z ? 5 into (2), we have y ? ?1.
Substituting y ? ?1 and z ? 5 into (1), we have x
? 3.
\ The unique solution of the system of linear
equations is x ? 3, y ? ?1, z ? 5.
32
12.4 Solving Equations by Gaussian
Elimination
In Gaussian elimination, since the elementary
transformations involve the coefficients of the
linear system only, we may use matrices to
shorten the operations.
First we need to define the augmented matrix
Definition 12.2 Augment Matrix Given a system of
linear equations, the matrix formed by adding a
column of constant terms to the right hand side
of the coefficient matrix is called the augmented
matrix of the system of linear equations.
33
12.4 Solving Equations by Gaussian
Elimination
Similar to the system of equations, we can also
define the row echelon form for a matrix
Definition 12.3 Row Echelon Form for Matrices A
matrix is said to be in row echelon form if it
satisfies the following conditions 1. The first
non-zero element in each row is 1. 2. For each
row which contains non-zero elements, the number
of leading zeros must be fewer than the
number of leading zeros in the row directly
below it. 3. The rows in which all elements are
zero are placed below the rows that have
non-zero elements.
Given an augmented matrix, we can transform it
into row echelon form using any of the following
three elementary row operations 1.
interchanging the position of two rows, 2.
multiplying a row by a non-zero number, 3.
adding an arbitrary multiple of any row to
another row.
34
12.4 Solving Equations by Gaussian
Elimination
Example 12.8T
Solution
(a)
? The unique solution of the system of linear
equations is x ? ?1, y ? 5, z ? 2.
35
12.4 Solving Equations by Gaussian
Elimination
Example 12.8T
Solution
(b)
? ? ? ?
36
12.4 Solving Equations by Gaussian
Elimination
Example 12.8T
Solution
(b)
\ The unique solution of the system of linear
equations is x ? 1, y ? ?2, z ? 1.
37
12.4 Solving Equations by Gaussian
Elimination
In addition to solving linear systems with a
unique solution, we can also use Gaussian
elimination to determine whether the equations in
the system are inconsistent or redundant, and
thus determine the number of solutions.
38
12.4 Solving Equations by Gaussian
Elimination
Example 12.9T
Solution
(a)
From equation (3), we have 0 ? ?2, which is
impossible. Thus, the system of linear equations
has no solution.
39
12.4 Solving Equations by Gaussian
Elimination
Example 12.9T
Solution
(b)
Hence the last equation is redundant which means
the system has infinitely many solutions.
40
12.4 Solving Equations by Gaussian
Elimination
Example 12.9T
Solution
Let z ? t, where t can be any real number.
(b)
Substituting z ? t into (2), we have
Substituting z ? t and y ? 2 3t into (1), we
have
41
12.4 Solving Equations by Gaussian
Elimination
Remarks The solutions of the systems of linear
equations that are expressed in terms of free
variable(s) are known as general solutions of the
systems.
The form of general solutions may not be unique.
42
12.4 Solving Equations by Gaussian
Elimination
Example 12.10T
Solution
43
12.4 Solving Equations by Gaussian
Elimination
Example 12.10T
(a) Find the values of a and b such that the
system of linear equations (E) has a unique
solution, and solve the system in cases where (E)
has solution(s).
Solution
(a) If the system of linear equations has a
unique solution, then A ? 0.
\ a 11 ? 0
Hence the conditions for (E) to have a unique
solution are a ? 11 and b can be any real number.
By Cramers rule,
\ The unique solution of the system is
44
12.4 Solving Equations by Gaussian
Elimination
Example 12.10T
(b) Find the values of a and b such that the
system of linear equations (E) has infinitely
many solutions, and solve the system in cases
where (E) has solution(s).
Solution
(b) If the system of linear equations does not
have a unique solution, then A ? 0, i.e.,
a ? 11.
Using Gaussian elimination,
Also if the system has infinitely many solutions,
we need b 7 ? 0.
Hence the conditions for (E) to have infinitely
many solutions are a ? 11 and b ? ?7.
45
12.4 Solving Equations by Gaussian
Elimination
Example 12.10T
(b) Find the values of a and b such that the
system of linear equations (E) has infinitely
many solutions, and solve the system in cases
where (E) has solution(s).
Solution
(b)
\ The system of equations can be expressed as
Let z ? t, where t is any real number.
Substituting z ? t into (2), we have y ? 2t 4.
Substituting z ? t and y ? 2t 4 into (1), we
have x ? ?3t ? 1.
\ The required solution is x ? 3t 1, y ? 2t
4, z ? t, where t is any real number.
46
12.4 Solving Equations by Gaussian
Elimination
Example 12.10T
(c) Find the values of a and b such that the
system of linear equations (E) has no solution,
and solve the system in cases where (E) has
solution(s).
Solution
(c) From (a) and (b), if the system of linear
equations has no solution, then A ? 0 and b 7
? 0.
Hence the conditions for (E) to have no solution
are a ? 11 and b ? 7.
47
12.5 Homogeneous Systems of Linear
Equations
For a system of linear equations Ax ? b, if the
constants bis are all zero, then the system is
said to be homogeneous.
In the previous sections, all the linear system
of equations discussed are non-homogeneous.
For solving a system of linear equations, we
learnt that there are three possible situations
1. it has a unique solution 2. it has no
solution 3. it has infinitely many solutions.
48
12.5 Homogeneous Systems of Linear
Equations
Thus a homogeneous system always has a solution,
and we call this solution a zero solution or a
trivial solution.
Thus there are only two possibilities for the
solutions of homogeneous systems of linear
equations
1. the system has only a trivial solution 2. a
non-trivial solution (i.e., not all x, y and z
are zeros) also exists.
The nature of the solutions of a homogeneous
system can be determined by the following
theorem
Theorem 12.5 If the number of unknowns in a
homogeneous system equals the number of
equations, then it has a non-trivial solution if
and only if the coefficient matrix is singular.
49
12.5 Homogeneous Systems of Linear
Equations
Proof If part Consider the linear system Ax ?
0.
If A is singular, then A ? 0. Thus, the
system does not have a unique solution.
? The system either has no solution, or has
infinitely many solutions.
Since the linear system has a trivial solution,
it is not possible for the system to have no
solution.
? The system must have infinitely many
solutions.
? The system must have non-trivial solutions.
Only if part We try to prove this by
contradiction.
Assume A is non-singular and the system has
non-trivial solutions.
? A is non-singular.
? A1 exists.
Then the system has a unique solution x ? A10 ?
0.
? The system has only trivial solution, which
contradicts our assumption.
? A must be singular.
50
12.5 Homogeneous Systems of Linear
Equations
Example 12.11T
Solution
(a) The determinant of the coefficient matrix
? 0
By Theorem 12.5, the system has non-trivial
solutions.
Using Gaussian elimination, we have
51
12.5 Homogeneous Systems of Linear
Equations
Example 12.11T
Solution
(a)
Let z ? t, where t can be any real number, then
we have x ? ?t and y ? t.
\ The required solution is x ? t, y ? t, z ? t,
where t can be any real number.
52
12.5 Homogeneous Systems of Linear
Equations
Example 12.11T
Solution
(b) The determinant of the coefficient matrix
? 2 ? 0
By Theorem 12.5, the system has a unique trivial
solution.
\ x ? 0, y ? 0, z ? 0.
53
12.5 Homogeneous Systems of Linear
Equations
Example 12.12T
Solution
(a) The system can be rewritten as
? The system of linear equations has non-trivial
solutions.
54
12.5 Homogeneous Systems of Linear
Equations
Example 12.12T
Given a system of linear equations (E)
, where k is a real
constant. (a) Find the values of k such that (E)
has non-trivial solutions. (b) Hence solve the
system of linear equations.
Solution
(a)
55
12.5 Homogeneous Systems of Linear
Equations
Example 12.12T
Given a system of linear equations (E)
, where k is a real
constant. (a) Find the values of k such that (E)
has non-trivial solutions. (b) Hence solve the
system of linear equations.
Solution
(b) For k ? 2,
56
12.5 Homogeneous Systems of Linear
Equations
Example 12.12T
Given a system of linear equations (E)
, where k is a real
constant. (a) Find the values of k such that (E)
has non-trivial solutions. (b) Hence solve the
system of linear equations.
Solution
(b)
\ z ? 0
Let y ? t, where t can be any real number, then
we have x ? ?t.
\ The required solution is x ? t, y ? t, z ? 0,
where t can be any real number.
57
12.5 Homogeneous Systems of Linear
Equations
Example 12.12T
Given a system of linear equations (E)
, where k is a real
constant. (a) Find the values of k such that (E)
has non-trivial solutions. (b) Hence solve the
system of linear equations.
Solution
(b)
For k ? 2,
58
12.5 Homogeneous Systems of Linear
Equations
Example 12.12T
Given a system of linear equations (E)
, where k is a real
constant. (a) Find the values of k such that (E)
has non-trivial solutions. (b) Hence solve the
system of linear equations.
Solution
(b)
Let z ? t, where t can be any real number, then
we have y ? ?2t, x ? ?2t.
\ The required solution is x ? 2t, y ? 2t, z ?
t, where t can be any real number.
59
12.5 Homogeneous Systems of Linear
Equations
Example 12.13T
Solution
The linear system can be rewritten as
.
Consider the determinant of the coefficient
matrix.
60
12.5 Homogeneous Systems of Linear
Equations
Example 12.13T
Solution
If the system has non-trivial solutions, then the
determinant D ? 0.
\ 5 l ? 0 or l2 38 ? 0
61
Chapter Summary
12.1 System of Linear Equations
62
Chapter Summary
12.2 Solving Equations by Inverses of Matrices
Consider a system of linear equations Ax ? b.
1. It has a unique solution, which is given by x
? A?1b, if and only is A ? 0.
2. It has either no solution or infinitely many
solutions if A ? 0.
63
Chapter Summary
12.3 Solving Equations by Cramers Rule
64
Chapter Summary
12.4 Solving Equations by Gaussian Elimination
A system of linear equations is said to be in row
echelon form if 1. the first non-zero term of
each row has a coefficient of 1.
2. in any two successive rows, for example, the
ith and (i 1)th rows, if the ith row does not
consist entirely of zero terms, then the number
of leading zeros in the (i 1)th row must be
greater than the number of leading zeros in the
ith row.
A system of equations can be transformed into the
row echelon form, without affecting its solution,
by any of the following elementary
transformations
1. interchanging the position of two equations
2. multiplying both sides of an equation by a
non-zero number
3. adding an arbitrary multiple of any equation
to another equation.
65
Chapter Summary
12.5 Homogeneous Systems of Linear Equations
66
Follow-up 12.1
12.2 Solving Equations by Inverses of
Matrices
Solution
\ The unique solution of the system of linear
equations is x ? 1, y ? 1.
67
Follow-up 12.2
12.2 Solving Equations by Inverses of
Matrices
Solution

• The unique solution of the system of
• linear equations is

68
Follow-up 12.3
12.2 Solving Equations by Inverses of
Matrices
Solution
\ The system does not have a unique solution.
(1) ? 2 (2) 0 ? 1, which is impossible.
Therefore, the system of linear equations has no
solution.
69
Follow-up 12.3
12.2 Solving Equations by Inverses of
Matrices
Solution
(b) Rewrite the system of equations as
.
\ The system does not have a unique solution.
(1) ? 3 9x 3y ? 3(3)
Since (2) and (3) are the same, we say that
equation (2) is redundant and the linear system
has only one equation 3x y ? 1.
Therefore, the system of linear equations has
infinitely many solutions.
70
Follow-up 12.4
12.3 Solving Equations by Cramers Rule
Solution
\ The unique solution of the system of linear
equations is
71
Follow-up 12.5
12.3 Solving Equations by Cramers Rule
Solution
\ The unique solution of the system of linear
equations is
72
12.3 Solving Equations by Cramers Rule
Follow-up 12.6
Solution
(a) The determinant of the coefficient matrix ?
? The system of linear equations has a unique
solution.

• D ? 0, i.e.,
• (b ? a)(c ? a)(c ? b) ? 0

\ a, b and c must be distinct.
73
12.3 Solving Equations by Cramers Rule
Follow-up 12.6
Solution
(b)
74
12.3 Solving Equations by Cramers Rule
Follow-up 12.6
Solution
(b)
75
12.3 Solving Equations by Cramers Rule
Follow-up 12.6
Solution
(b)
76
Follow-up 12.7
12.4 Solving Equations by Gaussian
Elimination
Solution
(a) Add the 3rd equation to the 2nd equation,
Interchange the 1st equation and the 3rd
equation,
Add the 2nd equation to the 1st equation,
which is the row echelon form of (E).
77
12.4 Solving Equations by Gaussian
Elimination
Follow-up 12.7
Solution
(b) From (3), we have z ? 5.
Substituting z ? 5 into (2), we have y ? ?1.
Substituting y ? ?1 and z ? 5 into (1), we have x
? 3.
\ The unique solution of the system of linear
equations is x ? 3, y ? ?1, z ? 5.
78
Follow-up 12.8
12.4 Solving Equations by Gaussian
Elimination
Solution
(a)
\ The unique solution of the system of linear
equations is x ? ?1, y ? ?2, z ? 3.
79
Follow-up 12.8
12.4 Solving Equations by Gaussian
Elimination
Solution
(b)
\ The unique solution of the system of linear
equations is x ? ?1, y ? ?5, z ? ?6.
80
Follow-up 12.9
12.4 Solving Equations by Gaussian
Elimination
Solution
(a)
From equation (3), we have 0 ? 12, which is
impossible. Thus, the system of linear equations
has no solution.
81
Follow-up 12.9
12.4 Solving Equations by Gaussian
Elimination
Solution
(b)
82
Follow-up 12.9
12.4 Solving Equations by Gaussian
Elimination
Solution
Hence the last equation is redundant which means
the system has infinitely many solutions.
(b)
Let z ? t, where t can be any real number.
Substituting z ? t into (2), we have
Substituting z ? t and y ? 2t 2 into (1), we
have
\ The unique solution is x ? 3t 2, y ? ?2t
2, z ? t, where t can be any real number.
83
Follow-up 12.10
12.4 Solving Equations by Gaussian
Elimination
Solution
84
Follow-up 12.10
12.4 Solving Equations by Gaussian
Elimination
(a) Find the values of m and n so that the system
of linear equations (E) has a unique solution,
and solve the system in cases where (E) has
solution(s).
Solution
(a) If the system of linear equations has a
unique solution, then A ? 0.
\ m 11 ? 0
Hence the condition for (E) to have a unique
solution are m ? 11 and n can be any real number.
By Cramers rule,
\ The unique solution of the system is
85
Follow-up 12.10
12.4 Solving Equations by Gaussian
Elimination
(b) Find the values of m and n so that the system
of linear equations (E) has infinitely many
solutions, and solve the system in cases where
(E) has solution(s).
Solution
(b) If the system of linear equations does not
have a unique solution, A ? 0, i.e., m ?
11. Using Gaussian elimination,
Also, if the system has infinitely many
solutions, we need 6 3n ? 0.
Hence the condition for (E) to have infinitely
many solutions are m ? 11 and n ? 2.
86
Follow-up 12.10
12.4 Solving Equations by Gaussian
Elimination
(b) Find the values of m and n so that the system
of linear equations (E) has infinitely many
solutions, and solve the system in cases where
(E) has solution(s).
Solution
\ The system of equations can be expressed as
(b)
Let z ? t, where t is any real number.
Substituting z ? t into (2), we have y ? 1 4t.
Substituting z ? t and y ? 1 4t into (1), we
have x ? 5t.
\ The required solution is x ? 5t, y ? 1 ? 4t, z
? t, where t is any real number.
87
Follow-up 12.10
12.4 Solving Equations by Gaussian
Elimination
(c) Find the values of m and n so that the system
of linear equations (E) has no solution, and
solve the system in cases where (E) has
solution(s).
Solution
(c) From (a) and (b), if the system of linear
equations has no solution, then A ? 0 and m
11 ? 0.
Hence the conditions for (E) to have no solution
are m ? 11 and n ? 2.
88
Follow-up 12.11
12.5 Homogeneous Systems of Linear
Equations
Solution
By Theorem 12.5, the system has non-trivial
solutions.
Using Gaussian elimination, we have
89
Follow-up 12.11
12.5 Homogeneous Systems of Linear
Equations
Solve the following system of linear equations
and determine whether they have trivial or
non-trivial solutions. (a) (b)
Solution
(a)
Let z ? t, where t can be any real number, then
we have x ? 0 and y ? ?t.
\ The required solution is x ? 0, y ? ?t, z ? t,
where t can be any real number.
90
Follow-up 12.11
12.5 Homogeneous Systems of Linear
Equations
Solve the following system of linear equations
and determine whether they have trivial or
non-trivial solutions. (a) (b)
Solution
By Theorem 12.5, the system has a unique trivial
solution.
\ x ? 0, y ? 0, z ? 0.
91
Follow-up 12.12
12.5 Homogeneous Systems of Linear
Equations
Solution
The system of linear equations has non-trivial
solutions.
92
Follow-up 12.12
12.5 Homogeneous Systems of Linear
Equations
Solution
(a)
93
Follow-up 12.12
12.5 Homogeneous Systems of Linear
Equations
Solution
For m ? ?1,
(b)
Let z ? t, where t can be any real number, then
we have y ? 0, x ? ?2t.
\ The required solution is x ? ?2t, y ? 0, z ?
t, where t can be any real number.
94
Follow-up 12.12
12.5 Homogeneous Systems of Linear
Equations
Solution
For m ? ?2,
(b)
Let z ? t, where t can be any real number, then
we have y ? ?t, x ? t.
\ The required solution is x ? t, y ? ? t, z ?
t, where t can be any real number.
95
Follow-up 12.13
12.5 Homogeneous Systems of Linear
Equations
Solution
The linear system can be rewritten as
Consider the determinant of the coefficient
matrix.
96
Follow-up 12.13
12.5 Homogeneous Systems of Linear
Equations
Consider the system of linear equations ()
Find the values of l such that () has
non-trivial solutions.
Solution
If the system has non-trivial solutions, then the
determinant D ? 0.