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

adjoint of matrix questions 3

   = [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
 

adjoint of matrix=

 
1 1 -1
-2 -6 3
-1 2 0
 







HTML Comment Box is loading comments...