Inversion Method Questions 3





In this page inversion method questions 3 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 3:

Solve the following linear equation by inversion method

x + y + z = 4

x - y + z = 2

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

To solve this we have to apply the formula X = A⁻¹ B.inversion method questions 3



|A|

=
 
1 1 1
1 -1 1
2 1 -1
 

  = 1

-1 1

1 -1

 -1

1 1

2 -1

+1

1 -1

2 1

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

      = 1 [0] - 1 [-3] +1 [1+2]

      = 0 + 3 +1 [3]

      = 0 + 3 + 3

      = 6

|A| = 6 ≠ 0

Since A is a non singular matrix. A⁻¹ exists.  inversion method questions 3

minor of 1

=
-1 1
1 -1

   = [1-1]

   = (0)

   = 0

Cofactor of 1

   =  + (0)

   =    0

minor of 1

=
1 1
2 -1

   = [-1-2]

   = (-3)

   = -3

Cofactor of 1

   =  - (-3)

   =    3

minor of 1

=
1 -1
2 1

   = [1-(-2)]

   = (1+2)

   = 3

Cofactor of 1

   =  + (3)

   =    3

minor of 1

=
1 1
1 -1

   = [-1-1]

   = (-2)

   = -2

Cofactor of 1

   =  - (-2)

   =    2

minor of -1

=
1 1
2 -1

   = [-1-2]

   = (-3)

   = -3

Cofactor of -1

   =  + (-3)

   =    -3

minor of 1

=
1 1
2 1

   = [1-2]

   = (-1)

   = -1

Cofactor of 1

   =  - (-1)

   =    1

minor of 2

=
1 1
-1 1

   = [1-(-1)]

   = (1+1)

   = 2

Cofactor of 2

   =  + (2)

   =    2

minor of 1

=
1 1
1 1

   = [1-1]

   = (0)

   = 0

Cofactor of 1

   =  - (0)

   =    0

minor of -1

=
1 1
1 -1

inversion method questions 3

   = [-1-1]

   = (-2)

   = -2

Cofactor of -1

   =  + (-2)

   =    -2

co-factor matrix =

 
0 3 3
2 -3 1
2 0 -2
 

adjoint of matrix=

 
0 2 2
3 -3 0
3 1 -2
 

            A⁻¹ = 1/6

 
0 2 2
3 -3 0
3 1 -2
 
 
x
y
z
 
 

  = 1/6

 
0 2 2
3 -3 0
3 1 -2
 
 
4
2
1
 
 


  = 1/6


 
0 2 2
 
x
 
4
2
1
 
 
3 -3 0
 
x
 
4
2
1
 
 
3 1 -2
 
x
 
4
2
1
 
 
x
y
z
 
 

=1/6

 
(0+4+2)
(12-6+0)
(12+2-2)
 
 

=1/6

 
(6)
(6)
(12)
 
 
 
x
y
z
 
 
=
 
1
1
2
 
 

Solution:

x = 1

y = 1

z = 2







Inversion Method Question3 to Inversion Method
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