## Adjoint of Matrix Questions 2

In this page adjoint of matrix questions 2 we are going to see solution of question 1 based on the topic ad-joint of matrix.

Question 2

Find the ad-joint of the following matrix

 1 2 3 1 1 1 2 3 4

Solution:

minor of 1

=
 1 1 -1 2

= [4-3]

= 1

Cofactor of 1

=  + (1)

=   1

minor of 2

=
 1 1 2 4

= [4-2]

= 2

Cofactor of 2

=  - (2)

=   -2

minor of 3

=
 1 1 2 3

= [3-2]

= 1

Cofactor of 3

=  + (1)

=   1

minor of 1

=
 2 3 3 4

= [8-9]

= -1

Cofactor of 1

=  - (-1)

=   1

minor of 1

=
 1 3 2 4

= [4-6]

= -2

Cofactor of 1

=  + (-2)

=   -2

minor of 1

=
 1 2 2 3

= [3-4]

= -1

Cofactor of 1

=  - (-1)

=   1

minor of 2

=
 2 3 1 1

= [2-3]

= -1

Cofactor of 2

=  + (-1)

=   -1

minor of 3

=
 1 3 1 1

= [1-3]

= -2

Cofactor of 3

=  - (-2)

=   2

minor of 4

=
 1 2 1 1

= [1-2]

= -1

Cofactor of 4

=  + (-1)

=   -1

co-factor matrix =

 1 -2 1 1 -2 1 -1 2 -1

 1 1 -1 -2 -2 2 1 1 -1 1. Click on the HTML link code below.

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