Inversion Method Questions 1





In this page inversion method questions 1 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 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
 
 

   = 2

1 1

-1 2

 - 1

 
1 1

1 2
 

+ 1

 
1 1

1 -1
 

|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.  inversion method questions 1

minor of 2

=
1 1
-1 2

inversion method questions 1

   = [2-(-1)]

   = (2+1)

   = 3

Cofactor of 2

   =  + (3)

   =    3

minor of 1

=
1 1
1 2

   = [2-1]

   = 1

Cofactor of 1

   =  -(1)

   =  -1

minor of 1

=
1 1
1 -1

   = [-1-1]

   = -2

Cofactor of 1

   =  + (-2)

   =  -2

minor of 1

=
1 1
-1 2

   = [2-(-1)]

   = [2+1]

   = 3

Cofactor of 1

   =  - (3)

   =  -3

minor of 1

=
2 1
1 2

   = [4-1]

   = 3

   = 3

Cofactor of 1

   =  + (3)

   =  3

minor of 1

=
2 1
1 -1

   = [-2-1]

   = -3

   = -3

Cofactor of 1

   =  - (-3)

   =  3

minor of 1

=
1 1
1 1

   = [1-1]

   = 0

Cofactor of 1

   =  + (0)

   =  0

minor of -1

=
2 1
1 1

   = [2-1]

   = 1

Cofactor of -1

   =  - (1)

   =  -1

minor of 2

=
2 1
1 1

   = [2-1]

   = 1

Cofactor of 2

   =  + (1)

   =  1

co-factor matrix =

 
3 -1 -2
-3 3 3
0 -1 1
 

adjoint of matrix=

 
3 -3 0
-1 3 -1
-2 3 1
 

            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
 
 


  = 1/3


 
3 -3 0
 
x
 
5
4
1
 
 
-1 3 -1
 
x
 
5
4
1
 
 


 
-2 3 1
 
x
 
5
4
1
 
 
 
x
y
z
 
 

=1/3

 
(15-12+0)
(-5+12-1)
(-10+12+1)
 
 

=1/3

 
(3)
(6)
(3)
 
 
 
x
y
z
 
 
=
 
1
2
1
 
 

Solution:

x = 1

y = 2

z = 1    inversion method questions 1







Inversion Method Question1 to Inversion Method
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