Inverse of Matrix Questions 3





In this page inverse of matrix questions 3 we are going to see solution of question 3 in the topic inverse of matrix.

Question 3

Find the inverse of the following matrix

 
6 2 3
3 1 1
10 3 4
 


Solution:

|A|

 = 6

 
1 1

3 4
 

 -2

 
3 1

10 4
 

+3

 
3 1

10 3
 

|A| = 6 [4-3] - 2 [12-10] + 3 [9-10]

      = 6 [1] - 2 [2] + 3 [-1]

      = 6 - 4 - 3

      = 6 - 7

      = -1

|A| = -1 ≠ 0

Since A is a non singular matrix. A⁻¹ exists.

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

inverse of matrix questions 3

   = [8-9]

   = (-1)

   = -1

Cofactor of 3

   =  - (-1)

   =  1

minor of 1

=
6 3
10 4

inverse of matrix questions 3

   = [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

co-factor matrix =

 
1 -2 -1
1 -6 2
-1 3 0
 

adjoint of matrix=

 
1 1 -1
-2 -6 3
-1 2 0
 

          A⁻¹ = 1/1

 
-1 -1 1
2 6 -3
1 -2 0
 









Questions



Solution


1) Find the inverse of the following matrix

 
2 1 1
1 1 1
1 -1 2
 

Solution

2) Find the inverse of the following matrix

 
1 2 1
2 -1 2
1 1 -2
 

Solution

4) Find the inverse of the following matrix

 
2 5 7
1 1 1
2 1 -1
 

Solution

5) Find the inverse of the following matrix

 
3 1 -1
2 -1 2
2 1 -2
 

Solution







Inverse of Matrix Question3 to Inverse of a Matrix
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