Solving a System of Equations Using Matrices
Solving a 2  2 system of linear equations by using the inverse matrix method
A system of linear equations can be solved by using our knowledge of inverse matrices.
The steps to follow are:
Express the linear system of equations as a matrix equation.
Determine the inverse of the coefficient matrix.
Multiply both sides of the matrix equation by the inverse matrix. In order to multiply
the matrices on the right side of the equation, the inverse matrix must appear in front
of the answer matrix.(the number of columns in the first matrix must equal the
number of rows in the second matrix).
Complete the multiplication.
 1 0  x   c1 
     where c1 and c2 are the solutions.
The solution will appear as: 
 0 1  y   c2 
Examples: Solve the following system of linear equations by using the inverse matrix
method:
2 x  9 y  1
1. 

4 x  y  15 
Solution:
If
 2 9  x    1

    
 4 1  y   15 
 2 9

A  
 4 1
 1

A 1    34
 4

  34
then
9 

 34 
2 

 34 
This is the matrix equation that represents the system.
A  2  36
A  34
 1

A 1   34
 4

 34
9 

34 
2

34 
This is the determinant and the inverse of the coefficient matrix.
 1

 34
 4

 34
9 
 1
 2 9  x  
34 
    34
 2  4 1  y   4


34 
 34
  2 36


 34 34
 8  8

 34 34
 34

 34
 0

 34
9 

34   1
 2  15 

34 
9 9 
 1 135 
  x  


34 34     34 34 
36  2  y    4  30 





34 34 
34 
 34
0
 136 
 x  

34     34 
34  y    34 



34 
 34 
 1 0  x   4 

    
 0 1  y    1
The common point or solution is (4, -1).
This is the result of multiplying the matrix equation by the inverse of the coefficient
matrix.
3x  6 y  45 
2. 

9 x  5 y  8
 3  6  x   45 
    
Solution: 
9

5

 y    8 
A  15  54
3  6
 then
If A  
A  39
9  5
5

1
A   39
9

 39
5

 39
9

 39
6

39 
3

39 
6
5
 3  6  x  
39 
    39
3  9  5  y    9


39 
 39
6

39  45 
3   8 

39 
  15 54


 39 39
  27  27

 39 39
 39

 39
 0

 39
30  30 
  225  48 


 x  

39 39     39
39 
54  15  y    405  24 





39 39 
39 
 39
0
  273 
 x  

39     39 
39  y    429 



39 
 39 
 1 0  x    7 

   

 0 1  y    11
The common point or solution is (-7, -11).
In the next example, the products will be written over the common denominator
instead of being written as two separate fractions.
4 x  y  13 
3. 

 6 x  5 y  37
1  x    13 
 4
   

Solution: 
  6  5  y   37 
 5

A 1    14
 6
  14
 5

 14
  6
 14
1 

 14 
4 

 14 
1 
 5
 4

1
x



14 
 14




 4   6  5  y    6


14 
 14
 20  6

 14
  24  24
 14
1 
 4

If A  
  6  5
then
 5

A 1   14
  6
 14
1 

14 
4

14 
1 

14   13 
 4  37 

14 
5  5 
  65  37 
 x  

14    
14

 6  20  y   78  148 



14 
14


A  20  6
A  14
 14 0 
  28 

 x  

 14 14     14 
 0 14  y    70 
 14 14 
 14 
 1 0  x    2 

    
 0 1  y    5 
The common point or solution is (-2, -5).
3x  y  11
4. 

x  2 y  8 
Solution:
 3  1 x    11

   

 1 2  y   8 
 3  1

If A  
1 2 
then
A  6  1
A 7
 2 1


A 1   7 7 
 1 3 


 7 7
 2 1
 2 1

 3  1 x  
  11
 7 7 
    7 7 

  1 3  1 2  y    1 3  8 




 7 7
 7 7
 6 1

 7
 33

 7
7

7
0

7
2 2
  22  8 
 x  

7     7 
1  6  y   11  24 



7 
 7 
0
  14 
 x  

 1 0  x    2 
7     7 

    
7  y   35 
0
1

 y   5 



7
7

 solution is (-2, 5)
The common point or
Exercises: Solve the following systems of linear equations by using the inverse matrix
method:
 5 x  3 y  21 
1. 

 2 x  7 y  21
2 x  3 y  48
2. 

3x  2 y  42
2 x  6 y  3
3. 

4 x  3 y  5
 x  y  1 
4. 

 4 x  2 y  8
Answers:
Solving systems of linear equations using the inverse matrix method
 5 x  3 y  21 
1. 

 2 x  7 y  21
If
  5 3

A  
 2 7
then A  35  6
A  29
 7

A 1    29
 2

  29
3 
7


 29   A 1   29
5 
2


 29 
 29
3 

29 
5 

29 
  5 3  x   21 

   

  2 7  y    21
7

 29
2

 29
3
7
  5 3  x  
29 
    29
5   2 7  y    2


29 
 29
 35  6

 29
 10  10

 29
 29

 29
 0

 29
3

29  21 
5   21

29 
 21  21 
  147  63 
 x  

29    
29

 6  35  y    42  105 



29 
29


0 
  210 
 x  

29     29 
29  y    147 



29 
 29 
 1 0  x    7.24 

   

 0 1  y    5.07 
2 x  3 y  48
2. 

3x  2 y  42
 2

1
A  5
3

5
 2 3
 then A  4  9
If A  
 3 2
A  5
3
2


 5   A 1   5
2 
 3


5
 5
3 

5 
2

5 
 2 3  x   48 

    
 3 2  y   42 
2 3 
2 3 

 2 3  x  

5 
5
5  48 
 5






 
 
 3  2  3 2  y   3  2  42 




5 
5 
 5
 5
49 66
  96  126 

 x  

5    
5
 5

 
 6  6 9  4  y   144  84 




5 
5
 5


5

5
0

5
0
 30 
 x   
5     5 
5  y   60 

 
5
 5 
 1 0  x   6 

    
 0 1  y  12 
2 x  6 y  3
3. 

4 x  3 y  5
 3 6 


A 1   18 18 
4 2 


 18 18 
 2  6
 then A  6  24
If A  
 4  3
A  18
 2  6  x   3 

    
 4  3  y   5 
 3 6 
 3 6 

 2  6  x  
 3 
 18 18 
    18 18  
  4 2  4  3  y    4 2  5 




 18 18 
 18 18 
  6  24 18  18 
  9  30 

 x  

18     18 
 18
24  6  y    12  10 
 88




18 
 18
 18 
 18 0 
 21 

 x  

 18 18     18 
 0 18  y    2 




 18 18 
 18 
 1 0  x   1.16 


   
0
1
y

.
1
1

  

 x  y  1 
4. 

 4 x  2 y  8
2

A 1   2
 4
2
 1 1
 then A  2  4
If A  
  4 2
A 2
1

2 
1

2 
  1 1   x  1

     
  4 2   y  8
2

2
 4
2
1
2
  1 1  x  
2 
    2
 1   4 2  y   4


2 
2
1

2  1 
 1  8 

2 
24

 2
  4  4
 2
2

2
 0
2
2  2 
 2  8 
 x  

2     2 
4  2  y   4  8 



2 
 2 
0
6
 x  

2     2 
2  y    4 



2
 2 
 1 0  x    3 

    
 0 1  y    2 