## 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  - 2 [-6] +1 

= 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

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

 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 