## Adjoint of Matrix Questions 5

In this page adjoint of matrix questions 5 we are going to see solution of question 5 based on the topic ad-joint of matrix.

Question 5

Find the ad-joint of the following matrix

 4 2 1 6 3 4 2 1 0

Solution:

minor of 4

=
 3 4 1 0

= [0-4]

= (-4)

= -4

Cofactor of 4

=  + (-4)

=    -4

minor of 2

=
 6 4 2 0

= [0-8]

= (-8)

= -8

Cofactor of 2

=  - (-8)

=    8

minor of 1

=
 6 3 2 1

= [6-6]

= (0)

= 0

Cofactor of 1

=  + (0)

=    0

minor of 6

=
 2 1 1 0

= [0-1]

= (-1)

= -1

Cofactor of 6

=  - (-1)

=    1

minor of 3

=
 4 1 2 0

= [0-2]

= (-2)

= -2

Cofactor of 3

=  + (-2)

=    -2

minor of 4

=
 4 1 2 0

= [4-4]

= (0)

= 0

Cofactor of 4

=  - (0)

=    0

minor of 2

=
 2 1 3 4

= [8-3]

= (5)

= 5

Cofactor of 2

=  + 5

minor of 1

=
 4 1 6 4

= [16-6]

= 10

Cofactor of 1

=  -(10)

= -10

minor of 0

=
 4 2 6 3

= [12-12]

= 0

Cofactor of 0

=  +(0)

= 0

co-factor matrix =

 -4 8 0 1 -2 0 5 -10 0

 -4 1 5 8 -2 -10 0 0 0 Adjoint of Matrix Question5 to Minor of a Matrix

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