## 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

 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