USING MATRIX INVERSE TO SOLVE A SYSTEM OF 3 LINEAR EQUATIONS

Example 1 :

Solve the following linear equation by inversion method

2x + y + z = 5

x + y + z = 4

x - y + 2z = 1

Solution :

First we have to write the given equation in the form

AX = B

Here X represents the unknown variables. A represent coefficient of the variables and B represents constants.



 
2 1 1
1 1 1
1 -1 2
 
 
x
y
z
 
 
=
 
5
4
1
 
 

To solve this we have to apply the formula X = A⁻¹ B

|A|

=
 
2 1 1
1 1 1
1 -1 2
 
 

|A|  =  2 [2-(-1)] - 1 [2-1] +1 [-1-1]

      =  2 [2+1] - 1 [1] +1 [-2]

      =  2 [3] - 1 -2

      =  6 - 3

      =  3

|A| = 3 ≠ 0

Since A is a non singular matrix. A⁻¹ exists. 

A⁻¹  =  1/3

 
3 -3 0
-1 3 -1
-2 3 1
 
 
x
y
z
 
 

  = 1/3

 
3 -3 0
-1 3 -1
-2 3 1
 
 
5
4
1
 
 

x  =  15-12+0

x  =  3

y  =  -5+12-1

y  =  6

z  =  -10+12+1

y  =  3

x  =  3, y  =  6 and z  =  3

Example 2 :

Solve the following linear equation by inversion method

x + 2y + z = 7

2x - y + 2z = 4

x + y - 2z = -1

Solution :



 
1 2 1
2 -1 2
1 1 -2
 
 
x
y
z
 
 
=
 
7
4
-1
 
 

To solve this we have to apply the formula X = A⁻¹ B

|A|

=
 
1 2 1
2 -1 2
1 1 -2
 
 

|A|  =  2 [2-2] -2[-4-2]+1[2+1]

=  2(0) - 2(-6) + 1(3)

=  0+12+3

|A|  =  15 ≠ 0

Since A is a non singular matrix. A⁻¹ exists. 

 
x
y
z
 
 

  = 1/15

 
0 5 5
6 -3 0
3 1 -5
 
 
7
4
-1
 
 

x  =  15/15

x  =  1

y  =  30/15

y  =  2

z  =  30/15

z  =  2

Example 3 :

Solve the following linear equation by inversion method

2x + y + z = 5

x + y + z = 4

x - y + 2z = 1 

Solution :



 
2 1 1
1 1 1
1 -1 2
 
 
x
y
z
 
 
=
 
5
4
1
 
 

To solve this we have to apply the formula X = A⁻¹ B

|A|

=
 
2 1 1
1 1 1
1 -1 2
 
 

  =  2(2+1)+1(2-1)+1(-1-1)

  =  2(3)-1-2

  =  3

|A|  =  3  ≠  0

Since A is a non singular matrix. A⁻¹ exists

 
x
y
z
 
 

  = 1/3

 
3 -3 0
-1 3 -1
-2 3 1
 
 
5
4
1
 
 

x  =  3/3

x  =  1

y  =  6/3

y  =  2

z  =  30/15

z  =  2

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