Adjoint of Matrix Questions 1





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

Question 1

Find the ad-joint of the following matrix

 
2 1 1
1 1 1
1 -1 2
 


Solution:

minor of 2

=
1 1
-1 2

   = [2-(-1)]

   = (2+1)

   = 3

Cofactor of 2

   =  + (3)

   =    3

minor of 1

=
1 1
1 2

   = [2-1]

   = 1

Cofactor of 1

   =  -(1)

   =  -1

minor of 1

=
1 1
1 -1

   = [-1-1]

   = -2

Cofactor of 1

   =  + (-2)

   =  -2

minor of 1

=
1 1
-1 2

   = [2-(-1)]

   = [2+1]

   = 3

   = 3

Cofactor of 1

   =  - (3)

   =  -3

minor of 1

=
2 1
1 2

   = [4-1]

   = 3

   = 3

Cofactor of 1

   =  + (3)

   =  3

minor of 1

=
2 1
1 -1

   = [-2-1]

   = -3

   = -3

Cofactor of 1

   =  - (-3)

   =  3

minor of 1

=
1 1
1 1

   = [1-1]

   = 0

Cofactor of 1

   =  + (0)

   =  0

minor of -1

=
2 1
1 1

   = [2-1]

   = 1

Cofactor of -1

   =  - (1)

   =  -1

minor of 2

=
2 1
1 1

adjoint of matrix questions 1 adjoint of matrix questions 1

   = [2-1]

   = 1

Cofactor of 2

   =  + (1)

   =  1

co-factor matrix =

 
3 -1 -2
-3 3 3
0 -1 1
 

adjoint of matrix=

 
3 -3 0
-1 3 -1
-2 3 1
 






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