Page 31 - mathematics

* We can write the previous system of equations in the Example
general form as:
Solve the following linear system of equations using
AX  B inverse matrix method:

where: x1  2x 2  2x 3  4
2x1  5x 2  x 3  7
a11 a12  a1n  x1  b1  4x1  5x 2  3x 3  5
a21    b2 
A   a22  a2n  , X   x2  , B   
       
Solution
an1 an 2  ann  x n  bn 
  

* To solve this system, multiply both sides by A 1 we get: 1 2 2 1 10 4 8
A  2 5 1 14  5 3
A 1A X  A 1B A 1   2 3 1 
X  A 1B 4 5 3
10

X  A 1B

x 1  1 10 4 8  4
 2  14 5
 x 3    2 3 3  7 
   
 x 10 1  5

 1 40  28  40
14 8  35  15
40  21  5 

1 28 2
14 42   
    3 

56   4 

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