In this page inverse of matrix questions 1 we are going to see solution of question 1 in the topic inverse of matrix.
Question 1
Find the inverse of the following matrix

Solution:
= 2 

 1 

+ 1 

A = 2 [2(1)]  1 [21] +1 [11]
= 2 [2+1]  1 [1] +1 [2]
= 2 [3]  1 2
= 6  3
= 3
A = 3 ≠ 0
Since A is a non singular matrix. A⁻¹ exists.
minor of 2 


= [2(1)] = (2+1) = 3  
Cofactor of 2 
= + (3) = 3  
minor of 1 


= [21] = 1  
Cofactor of 1 
= (1) = 1  
minor of 1 


= [11] = 2  
Cofactor of 1 
= + (2) = 2  
minor of 1 


= [2(1)] = [2+1] = 3  
Cofactor of 1 
=  (3) = 3  
minor of 1 


= [41] = 3 = 3  
Cofactor of 1 
= + (3) = 3  
minor of 1 


= [21] = 3 = 3  
Cofactor of 1 
=  (3) = 3  
minor of 1 


= [11] = 0  
Cofactor of 1 
= + (0) = 0  
minor of 1 


= [21] = 1  
Cofactor of 1 
=  (1) = 1  
minor of 2 


= [21] = 1  
Cofactor of 2 
= + (1) = 1 
inverse of matrix questions 1
cofactor matrix = 


adjoint of matrix= 


A⁻¹ = 1/3 

Questions 
Solution  
2) Find the inverse of the following matrix 


3) Find the inverse of the following matrix 


4) Find the inverse of the following matrix 


5) Find the inverse of the following matrix 


inverse of matrix questions 1 