## Inversion Method Questions 4

In this page inversion method questions 4 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 4:

Solve the following linear equation by inversion method

2x + 5y + 7z = 52

x + y + z = 9

2x + y - z = 0

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 5 7 1 1 1 2 1 -1

 x y z

=

 52 9 0

|A|

= 2

 1 1 1 -1

-5

 1 1 2 -1

+7

 1 1 2 1

|A| = 2 [-1-1] - 5 [-1-2] + 7 [1-2]

= 2 [-2] - 5 [-3] + 7 [-1]

= -4 + 15 - 7

= -11 + 15

= 4

|A| = 4 ≠ 0

Since A is a non singular matrix. A⁻¹ exists.inversion method questions 4

minor of 2

=
 1 1 1 -1

inversion method questions 4

= [-1-1]

= (-2)

= -2

Cofactor of 2

=  + (-2)

=    -2

minor of 5

=
 1 1 2 -1

= [-1-2]

= -3

Cofactor of 5

=  - (-3)

=    3

minor of 7

=
 1 1 2 1

= [1-2]

= -1

Cofactor of 7

=  + (-1)

=    -1

minor of 1

=
 5 7 1 -1

= [-5-7]

= -12

Cofactor of 1

=  - (-12)

=    12

minor of 1

=
 2 7 2 -1

= [-2-14]

= -16

Cofactor of 1

=  + (-16)

=   -16

minor of 1

=
 2 5 2 1

= [2-10]

= -8

Cofactor of 1

=  - (-8)

=   8

minor of 2

=
 5 7 1 1

= [5-7]

= -2

Cofactor of 2

=  + (-2)

=   -2

minor of 1

=
 2 7 1 1

= [2-7]

= -5

Cofactor of 1

=  - (-5)

=   5

minor of -1

=
 2 5 1 1

inversion method questions 4

= [2-5]

= -3

Cofactor of -1

=  + (-3)

=   -3

co-factor matrix =

 -2 3 -1 12 -16 8 -2 5 -3

 -2 12 -2 3 -16 5 -1 8 -3

A⁻¹ = 1/4

 -2 12 -2 3 -16 5 -1 8 -3

 x y z

= 1/4

 -2 12 -2 3 -16 5 -1 8 -3

 52 9 0

= 1/4

 -2 12 -2

x

 52 9 0

 3 -16 5

x

 52 9 0

 -1 8 -3

x

 52 9 0

 x y z

=1/4

 (-104+108+0) (156-144+0) (-52+72+0)

=1/4

 (4) (12) (20)

 x y z

 1 3 5

Solution:

x = 1

y = 3

z = 5 Inversion Method Question4 to Inversion Method 