## 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

= [2-1]

= 1

Cofactor of 2

=  + (1)

=  1

co-factor matrix =

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

 3 -3 0 -1 3 -1 -2 3 1 1. Click on the HTML link code below.

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