## Adjoint of Matrix Questions 3

In this page adjoint of matrix questions 3 we are going to see solution of question 1 based on the topic ad-joint of matrix.

Question 3

Find the ad-joint of the following matrix

 6 2 3 3 1 1 10 3 4

Solution:

minor of 6

=
 1 1 3 4

= [4-3]

= (1)

= 1

Cofactor of 6

=  + (1)

=    1

minor of 2

=
 3 1 10 4

= [12-10]

= (2)

= 2

Cofactor of 2

=  - (2)

=  -2

minor of 3

=
 3 1 10 3

= [9-10]

= (-1)

= -1

Cofactor of 3

=  + (-1)

=  -1

minor of 3

=
 2 3 3 4

= [8-9]

= (-1)

= -1

Cofactor of 3

=  - (-1)

=  1

minor of 1

=
 6 3 10 4

= [24-30]

= (-6)

= -6

Cofactor of 1

=  + (-6)

=  -6

minor of 1

=
 6 2 10 3

= [18-20]

= (-2)

= -2

Cofactor of 1

=  - (-2)

=  2

minor of 10

=
 2 3 1 1

= [2-3]

= (-1)

= -1

Cofactor of 10

=  + (-1)

=  -1

minor of 3

=
 6 3 3 1

= [6-9]

= (-3)

= -3

Cofactor of 3

=  - (-3)

=  3

minor of 4

=
 6 2 3 1

= [6-6]

= (0)

= 0

Cofactor of 4

=  + (0)

=  0              adjoint of matrix questions 3

co-factor matrix =

 1 -2 -1 1 -6 2 -1 3 0

 1 1 -1 -2 -6 3 -1 2 0

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