Inversion Method Questions 2





In this page inversion method questions 2 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 2:

Solve the following linear equation by inversion method

x + 2y + z = 7

2x - y + 2z = 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.



 
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

 
-1 2

1 -2
 

 -2

 
2 2

1 -2
 

+1

 
2 -1

1 1
 

|A| = 1 [2-2] - 2 [-4-2] +1 [2+1]

      = 1 [0] - 2 [-6] +1 [3]

      = 0 + 12 + 3

      = 15

|A| = 15 ≠ 0

Since A is a non singular matrix. A⁻¹ exists.   inversion method questions 2

minor of 1

=
-1 2
1 -2

inversion method questions 2

   = [2-2]

   = 0

Cofactor of 1

   =  + (0)

   =    0

minor of 2

=
2 2
1 -2

   = [-4-2]

   = -6

Cofactor of 2

   =  - (-6)

   =    6

minor of 1

=
2 -1
1 1

   = [2-(-1)]

   = [2+1]

   = 3

Cofactor of 1

   =  + (3)

   =    3

minor of 2

=
2 1
1 -2

   = [-4-1]

   = -5

Cofactor of 2

   =  - (-5)

   =    5

minor of -1

=
1 1
1 -2

   = [-2-1]

   = -3

Cofactor of -1

   =  + (-3)

   =    -3

minor of 2

=
1 2
1 1

   = [1-2]

   = -1

Cofactor of 2

   =  - (-1)

   =    1

minor of 1

=
2 1
-1 2

   = [4-(-1)]

   = [4+1]

   = 5

Cofactor of 1

   =  + (5)

   =    5

minor of 1

=
1 1
2 2

   = [2-2]

   = [0]

   = 0

Cofactor of 1

   =  - (0)

   =    0

minor of -2

=
1 2
2 -1

inversion method questions 2

   = [-1-4]

   = [-5]

   = -5

Cofactor of -2

   =  + (-5)

   =    -5

co-factor matrix =

 
0 6 3
5 -3 1
5 0 -5
 

adjoint of matrix=

 
0 5 5
6 -3 0
3 1 -5
 

          A⁻¹ = 1/15

 
0 5 5
6 -3 0
3 1 -5
 
 
x
y
z
 
 

  = 1/15

 
0 5 5
6 -3 0
3 1 -5
 
 
7
4
-1
 
 

 
0 5 5
 
x
 
7
4
-1
 

    = 1/15

 
6 -3 0
 
x
 
7
4
-1
 
 
3 1 -5
 
x
 
7
4
-1
 

 
x
y
z
 
 

    =1/15


 
(0+20-5)
(42-12-0)
(21+4+5)
 
 

 =1/15

 
(15)
(30)
(30)
 
 
 
x
y
z
 
 
=
 
1
2
2
 
 

Solution:

x = 1

y = 2

z = 2           







Inversion Method Question2 to Inversion Method
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