## Inversion Method Questions 1

In this page inversion method questions 1 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 1:

Solve the following linear equation by inversion method

2x + y + z = 5

x + y + z = 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.

 2 1 1 1 1 1 1 -1 2

 x y z

=

 5 4 1

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

|A|

=

 2 1 1 1 1 1 1 -1 2

= 2

 1 1 -1 2

- 1

 1 1 1 2

+ 1

 1 1 1 -1

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

= 2 [2+1] - 1  +1 [-2]

= 2  - 1 -2

= 6 - 3

= 3

|A| = 3 ≠ 0

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

minor of 2

=
 1 1 -1 2

inversion method questions 1

= [2-(-1)]

= (2+1)

= 3

Cofactor of 2

=  + (3)

=    3

minor of 1

=
 1 1 1 2

= [2-1]

= 1

Cofactor of 1

=  -(1)

=  -1

minor of 1

=
 1 1 1 -1

= [-1-1]

= -2

Cofactor of 1

=  + (-2)

=  -2

minor of 1

=
 1 1 -1 2

= [2-(-1)]

= [2+1]

= 3

Cofactor of 1

=  - (3)

=  -3

minor of 1

=
 2 1 1 2

= [4-1]

= 3

= 3

Cofactor of 1

=  + (3)

=  3

minor of 1

=
 2 1 1 -1

= [-2-1]

= -3

= -3

Cofactor of 1

=  - (-3)

=  3

minor of 1

=
 1 1 1 1

= [1-1]

= 0

Cofactor of 1

=  + (0)

=  0

minor of -1

=
 2 1 1 1

= [2-1]

= 1

Cofactor of -1

=  - (1)

=  -1

minor of 2

=
 2 1 1 1

= [2-1]

= 1

Cofactor of 2

=  + (1)

=  1

co-factor matrix =

 3 -1 -2 -3 3 3 0 -1 1

 3 -3 0 -1 3 -1 -2 3 1

A⁻¹ = 1/3

 3 -3 0 -1 3 -1 -2 3 1

 x y z

= 1/3

 3 -3 0 -1 3 -1 -2 3 1

 5 4 1

= 1/3

 3 -3 0

x

 5 4 1

 -1 3 -1

x

 5 4 1

 -2 3 1

x

 5 4 1

 x y z

=1/3

 (15-12+0) (-5+12-1) (-10+12+1)

=1/3

 (3) (6) (3)

 x y z

=

 1 2 1

Solution:

x = 1

y = 2

z = 1    inversion method questions 1 Inversion Method Question1 to Inversion Method 