Inversion Method Questions 5





In this page inversion method questions 5 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 5:

Solve the following linear equation by inversion method

3x + y - z = 2

2x - y + 2z = 6

2x + y - 2z = -2

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.



 
3 1 -1
2 -1 2
2 1 -2
 
 
x
y
z
 
 
=
 
2
6
-2
 
 

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

|A|

=
 
3 1 -1
2 -1 2
2 1 -2
 
 

|A|

 = 3

 
-1 2

1 -2
 

- 1

 
2 2

2 -2
 

-1

 
2 -1

2 1
 

|A| = 3 [2-2] - 1 [-4-4] - 1 [2-(-2)]

      = 3 [0] - 1 [-8] -1 [2+2]

      = 3 [0] - 1 [-8] -1 [4]

      = 0 + 8 - 4

      = 4

|A| = 4 ≠ 0

Since A is a non singular matrix. A⁻¹ exists.inversion method questions 5

minor of 4

=
-1 2
1 -2

   = [2-2]

   = 0

Cofactor of 4

   =  + (0)

   =    0

minor of 1

=
2 2
2 -2

   = [-4-4]

   = (-8)

   = -8

Cofactor of 1

   =  - (-8)

   =    8

minor of -1

=
2 -1
2 1

nversion method questions 5

   = [2-(-2)]

   = (2+2)

   = 4

Cofactor of -1

   =  + (4)

   =    4

minor of 2

=
1 -1
1 -2

   = [-2-(-1)]

   = (-2+1)

   = -1

   = -1

Cofactor of 2

   =  - (-1)

   =    1

minor of -1

=
3 -1
2 -2

   = [-6-(-2)]

   = (-6+2)

   = -4

   = -4

Cofactor of -1

   =  + (-4)

   =    -4

minor of 2

=
3 1
2 1

   = [3-2]

   = 1

Cofactor of 2

   =  - (1)

   =    -1

minor of 2

=
1 -1
-1 2

   = [2-1]

   = 1

Cofactor of 2

   =  + (1)

   =    1

minor of 1

=
3 -1
2 2

   = [6-(-2)]

   = [6+2]

   = 8

Cofactor of 1

   =  - (8)

   =    -8

minor of -2

=
3 1
2 -1

inversion method questions 5

   = [-3-2]

   = [-5]

   = -5

Cofactor of -2

   =  + (-5)

   =    -5

co-factor matrix =

 
0 8 4
1 -4 -1
5 0 -5
 

adjoint of matrix=

 
0 1 5
8 -4 0
4 -1 -5
 

          A⁻¹ = 1/4

 
0 1 5
8 -4 0
4 -1 -5
 
 
x
y
z
 
 

  = 1/4

 
0 1 5
8 -4 0
4 -1 -5
 
 
2
6
-2
 
 


  = 1/4


 
0 1 5
 
x
 
2
6
-2
 
 
8 -4 0
 
x
 
2
6
-2
 
 
4 -1 -5
 
x
 
2
6
-2
 
 
x
y
z
 
 

=1/4

 
(0+6-10)
(16-24+0)
(8-6+10)
 
 

=1/4

 
(-4)
(-8)
(12)
 
 
 
x
y
z
 
 
=
 
-1
-2
3
 
 

Solution:

x = -1

y = -2

z = 3          inversion method questions 5  inversion method questions 5







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