In this page inversion method questions 1 we are going to see how to solve the given linear equations using this particular method in matrices.
Question 1:
Solve the following linear equation by inversion method
2x + y + z = 5
x + y + z = 4
x  y + 2z = 1
Solution:
First we have to write the given equation in the form AX = B. Here X
represents the unknown variables. A represent coefficient of the
variables and B represents constants.



To solve this we have to apply the formula X = A⁻¹ B
A 

= 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. inversion method questions 1
minor of 2 


inversion method questions 1 
= [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 
cofactor matrix = 


adjoint of matrix= 


A⁻¹ = 1/3 


= 1/3 


= 1/3 
 

 

 


Solution:
x = 1
y = 2
z = 1 inversion method questions 1