Lecture 7

2.10 Systems of Linear Equations

Any system of linear equations can be represented in
matrix notation as shown below.
a11 x  a12 y  a13 z  b1

a21 x  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 
where A is known as the coefficient matrix.

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

The given system can be written as

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

AX  B  A1 AX  A1B
I . X  A1B
1
 
X  A1B   2 
  1
 

Method 2
When the coefficient matrix of the system is a square
matrix and the determinant of the coefficient matrix
is not equal to zero , the solution of the system is
obtained using the Cramer’s Rule given below.
Cramer’s Rule
The solution of the system of n equations in n
unknowns x1 , x2 ,........, xn given by,
a11 x1  a12 x2  .......... .  a1n xn  b1
a21 x1  a22 x2  .......... .  a2 n xn  b2
. . . .
. . . .

an1 x1  an 2 x2  .......... .  ann xn  bn

Ai
xi  i  1,2,......., n whenever
is equal to A for
A 0
where A is the determinant of the coefficient matrix
and Ai is the determinant obtained from the
coefficient matrix by replacing its ith column by the
column containing the numbers b1 , b2 ,........, bn .
Example

Using Cramer’s Rule solve the following system.

x  3y  z  8
2 x  y  3z  7
x yz2
1 3 1 
 
A  2 1 3 
The coefficient matrix,  1 1  1
 

8 3 1
7 1 3
A1 2 1 1 12
A  12 , then x    1
A 12 12

1 8 1
2 7 3
A2 1 2 1 24
y   2
A 12 12
1 3 8
2 1 7
A3 1 1 2 12
z   1
A 12 12
Method 3
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.
The process of reducing the augmented matrix to
row echelon form using a series of row
transformations is known as Gaussian
elimination.
Examples
1.Solve the following system using Gaussian
elimination.
2 x  3 y  4 z  10
x  2 y  z  1
3x  4 y  3

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

1 2 1 1 
 
  0 1 6 12 
 0 0  9  18
 

 1 2 1 1 
  1
  0 1  6  12  , R2   R2 , R3  R3
0 0 1  9
 2 
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
 , R  R2 , R3  R3  R4
9  12 
2
0 0 5
 
0 2   
 4 4 
1 1 3 5 
 
0 1  8 / 5  12 / 5  1 1
 , R  R , R  R3
4/3  9
4 4 3
0 0 1 2
 
0 2  2 
 1

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