In this page adjoint of matrix questions 3 we are going to see solution of question 1 based on the topic adjoint of matrix.
Question 3
Find the adjoint of the following matrix

Solution:
minor of 6 


= [43] = (1) = 1  
Cofactor of 6 
= + (1) = 1 
minor of 2 


= [1210] = (2) = 2  
Cofactor of 2 
=  (2) = 2 
minor of 3 


= [910] = (1) = 1  
Cofactor of 3 
= + (1) = 1 
minor of 3 


= [89] = (1) = 1  
Cofactor of 3 
=  (1) = 1 
minor of 1 


= [2430] = (6) = 6  
Cofactor of 1 
= + (6) = 6 
minor of 1 


= [1820] = (2) = 2  
Cofactor of 1 
=  (2) = 2 
minor of 10 


= [23] = (1) = 1  
Cofactor of 10 
= + (1) = 1 
minor of 3 


= [69] = (3) = 3  
Cofactor of 3 
=  (3) = 3 
minor of 4 


adjoint of matrix questions 3 
= [66] = (0) = 0  
Cofactor of 4 
= + (0) = 0 adjoint of matrix questions 3 
cofactor matrix = 


adjoint of matrix= 

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