2.

7 Systems of Linear Equations

Any system of linear equations can be represented by
matrix notation as shown below.
a11 x  a12 y  a13 z  b1
a21x  a22 y  a23 z  b2

a31 x  a32 y  a33 z  b3

In matrix notation this system can be written as
 a11 a12 a13  x   b1 
    
 a21 a22 a23  y    b2 
a      AX  B
 31 a32 a33  z   b3 

If B  O then the system is called non- homogeneous

otherwise homogeneous.

Solution of non-homogeneous systems

Method 1
This method is used for a system where
the number of linear equations = number of
unknowns and the inverse of coefficient matrix exists.
AX  B .................(1)
A1  (1)  A1 AX  A1B
I . X  A1B
X  A1B
i.e. A  0, the system has a unique solution.
Example
 2 3 4
 
A  1 2 1
Find the inverse of  4 6 5  . Hence solve the
 
following system of equations.

2x  3y  4z  4
x  2y  z  4
4 x  6 y  5 z  11

 4/ 3 3 5/3 
 
It can be shown that A1   1/ 3 2  2 / 3
 2/3  1 / 3 
 0

 x
   2 3 4 4
AX  B X   y 
A  1 2 1
  
B 4
where z ,  4 6 5
and 11
     
 1
1
 
X  A B   2
1
 

Method 2
Reducing to echelon form
In the system AX  B the matrix A is called the
coefficient matrix and coefficient matrix together
 a11 a12 a13 : b1 
 
with constants i.e.  21 22
a a a13 : b2 

is called the
 a13 a23 a33 : b3 
augmented matrix.
When a series of elementary row transformations are
applied on augmented matrix a special matrix which
is equivalent to the augmented matrix is obtained.
This matrix is called the echelon form of the matrix.
In echelon form, the number of zeros before the first
non zero element in each row is in increasing order
with each first non zero element is equal to 1.
Examples
1.Solve the following system reducing to the echelon
form.
2 x  3 y  4 z  10
x  2 y  z  1
3x  4 y  3

1 2  1  1

 2 3 4 10 
   2 
3 4 10  , R1  R2
 1 2  1  1 
3 4 0 3  34 0 3 
Consider   

 1 2  1  1
 
  0  1 6 12  , R2  R2  2 R1 , R3  R3  3R1
0  2 3 6 
 

1 2 1 1 
 
  0 1 6 12 
 0 0  9  18 
  , R3  R3  2 R2

1 2 1 1 
 
  0 1  6  12  1
0 0 1
 2  , R3  R3
9
 The reduced system is

x  2 y  z  1
y  6 z  12
z2

Solution is x  1, y  0, z  2

2. Obtain the echelon form of the matrix

3 2 1 3
 
1 1 3 5
2  4 1 2  . Hence solve the system
 
1 1 1 1 


3x  2 y  z  3
x  y  3z  5
2x  4 y  z  2
x yz 1
3 2 1 3 1 1 3 5
   
1 1 3 5 3 2 1 3
2  4 1   , R1  R2
2 2 4 1 2
   
1 1 1   1 
 1 1 1 1

1 1 3 5 
 
 0 5  8  12 
 , R2  R2  3R1 , R3  R3  2 R1 , R4  R4  R1
0  2 5 8 
 
0 2  4  4 
 

1 1 3 5 
 
 0 1  8 / 5  12 / 5  1
  , R2  R2 , R3  R3  R4
0 0 9  12 5
 
0 2 4  4 


1 1 3 5 
 
 0 1  8 / 5  12 / 5  1 1
  , R4  R4 , R3  R3
0 0 1 4/3 2 9
 
0 1 2  2 


1 1 3 5 
 
 0 1  8 / 5  12 / 5 
 , R4  R4  R2
0 0 1 4/3 
 
0 0  2 / 5 2 / 5 
 

1 1 3 5 
 
 0 1  8 / 5  12 / 5  5
  , R4  R4
0 0 1 4/3 2
 
0 0 1 
 1 

1 1 3 5 
 
 0 1  8 / 5  12 / 5 
 , R4  R4  R3
0 0 1 4/3 
 
0 0 7 / 3 
 0
 1 1 3 5 
 
 0 1  8 / 5  12 / 5  3
  , R4  R4
0 0 1 4/3 7
 
0 0 
 0 1 

The reduced system is

x  y  3z  5
8  12
y z
5 5
4
z
3
0 1
Last equation is a contradiction.
Therefore the given system has no solutions.

Rank of a Matrix
Number of non zero rows(columns) of the
row(column) echelon form of a matrix is defined as
the rank of a matrix.
Exercise
1. Solve the following system of equations
x  2 y  z  1
x  y  2z  3
4 x  y  5z  8
5 x  7 y  2 z  1

Augmented matrix of the system is

 1  2 1  1  1  2 1  1
   
1 1  2 3  0 3  3 4 
 4 1  5 8    0 9  9 12  , R2  R2  R1 , R3  R3  4 R1 , R4  R4  5R1
   
 5  7 2  1 0 3  3 4 
   

 1  2 1  1
 
 0 3  3 4  1
0 3  3 4  , R3  R3 , R4  R4  R2
 3
 
0 0 0 
 0

 1  2 1  1
 
0 3  3 4 
0 0  , R3  R3  R2
0 0
  
0 0 
 0 0 
  1  2 1  1
 4
0 1 1  1
 3  , R2  R2
0 0 0 0 3

0 0 0 0 


Therefore reduced system is

x  2 y  z  1
4
yz
3
Let z t where t is a parameter.

Then z t
1
x  (3t  5)
3

The solution of the system is

1
x  (3t  5)
3
1
y  (4  3t )
3
z t
Since t is a parameter the system has infinite
number of solutions.

2. Show that the following system of equations has a

unique solution unless a  2.

x  y  2 z  3
2x  3 y  4z  5
5x  az  b
 1  1 2  3
 
Augmented matrix = 2 3  4 5 
5 0 b 
 a

1 1 2 3 
 
0 5 8 11  , R  R  2R , R  R  5R
 0 5 (a  10) (b  15)  2 2 1 3 3 1

 

1 1 2 3 
 
 0 5  8 11 , R3  R3  R2
 0 0 (a  2) (b  4) 
 

1 1 2 3 
 8 11 
0 1  1
 5 5  , R2 
5
R2
 0 0 (a  2) (b  4) 
If a  2
 
 
1 1 2 3 
0 1  8 11 
 5 5 , R3 
1
R3
 b4 (a  2)
0 0 1 
 a2
Then reduced system is
x  y  2 z  3

8 11
y  z 
5 5

b4
z
a2

If a  2 the system has a unique solution.

Note

For a non- homogeneous system Anxn X nx1  Bnx1

If A  0 the system has a unique solution

( i.e. rank of A = rank of augmented matrix =n)

If A  0 the system has infinite no of solutions or no

solutions
 (i.e. rank of A  rank of the augmented matrix then
the system has no solutions

rank of A = rank of augmented matrix < n then

the system has infinite number of
solutions)

If the system has solutions we say it is consistent

otherwise inconsistent.

Example

For what values of  ,  the simultaneous equations

x  2y  z  8
2 x  y  3 z  13
3 x  4 y  z  
have (i) no solutions
(ii) a unique solution
(iii) infinitely many solutions

1 2 1 
 
A  2 1 3
Let 3 4  
 
 1 2 1
A  2 1 3  3  11
3 4 

If A  0 i.e. if  3  11  0 , 
11
the system has a
3

unique solution for any value of  .

If A  0 i.e. if 
11
the system has no solutions or
3

infinite number of solutions.

 
1 2 1 8 
Augmented matrix =  2 1 3 13 
 11 
3 4 
 3 

 
1 2 1 8 
 2 1 3 13 
 11 
3 4 
 3 

 
1 2 1 8 
0  3 1 3  , R2  R2  2 R1 , R3  R3  3R1
 2 
0  2   24 
 3 

1 2 1 8 
  2
0  3 1 3  , R3  R3 
3
R2
 0 0 0   22 
 
 1 2 1 8 
 1  1
0 1 1  , R2  R2
 3  3
0 0 0   22 

If   22

1 2 1 8
 1 
0 1 1
 3 
0 0 0 0 

Rank A = Rank (A B)=2 < 3

The given system has infinite number of solutions.
If   22

1 2 1 8
 1 
Augmented matrix 0 1 1
 3 
0 0 0 1 

Rank A  Rank (A B)
The given system has no solutions.
Thus we get that
11
(i) If   3 ,   22 then the system has no
solutions.
11
(ii) If   3 then the system has a unique
solution for any value of  .
11
(iii) If   3 ,   22 then system has infinite
number of solutions.

Solution of homogeneous systems

The system is of the form AX  O . Always X  O is a

solution of the system.
X  O is called the trivial solution. Therefore
homogeneous system is always consistent.
Examples
Solve the following systems
1. 2x  7 y  6z  0 2. 2 x  y  3z  0
3x  5 y  2 z  0 3x  2 y  z  0
4x  2 y  7z  0 x  4 y  5z  0

Note: For a homogeneous system,

If A 0 , the system has only the trivial solution.
If A 0 , the system has solutions other than the
trivial solution.
Summary
System of linear Equations ( Amxn X nx1  Bmx1 )

Non Homogeneous( B  o ) Homogeneous( B  o )

consistent inconsistent consistent

(has solutions) ( no solutions) always trivial solution exists

If m=n
If A  0 the system has a unique If A  0, only trivial solutionexists
solution

If A  0 the system has no If A 0, has solutions other than

solutions or infinitely many the trivial solutions.
solution

If m n
If Rank(A) = Rank(A B)=n If Rank(A) = Rank(A B)=n only
the system has a unique trivial solution exists.
Solution.

If Rank(A) = Rank(A B) <n If Rank(A) = Rank(A B) <n

Infinite no. of solutions solutions other than the
Exists. trivial solution exists
If Rank(A)  Rank(A B)
No solutions for the system

2. 2 x  7 y  6 z  0 2. 2 x  y  3z  0
3x  5 y  2 z  0
 3x  2 y  z  0

4x  2 y  7z  0 x  4 y  5z  0

Note: For a homogeneous system,

If A 0 , the system has only the trivial solution.
If A  0 , the system has solutions other than the
trivial solution.

Summary
System of linear Equations ( A mxn X nx1  Bmx1 )

Non Homogeneous( B  o )
Homogeneous( B  o )

consistent inconsistent
consistent
(has solutions) ( no solutions)
always trivial solution exists
If m=n
If A  0 the system has a unique If A 0

, only trivial solution exists

solution