# USING MATRIX INVERSE TO SOLVE A SYSTEM OF 3 LINEAR EQUATIONS

Example 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.

 2 1 1 1 1 1 1 -1 2

 x y z

=

 5 4 1

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

|A|

=

 2 1 1 1 1 1 1 -1 2 |A|  =  2 [2-(-1)] - 1 [2-1] +1 [-1-1]

=  2 [2+1] - 1  +1 [-2]

=  2  - 1 -2

=  6 - 3

=  3

|A| = 3 ≠ 0

Since A is a non singular matrix. A⁻¹ exists. A⁻¹  =  1/3

 3 -3 0 -1 3 -1 -2 3 1

 x y z

= 1/3

 3 -3 0 -1 3 -1 -2 3 1

 5 4 1

 x  =  15-12+0x  =  3 y  =  -5+12-1y  =  6 z  =  -10+12+1y  =  3

x  =  3, y  =  6 and z  =  3

Example 2 :

Solve the following linear equation by inversion method

x + 2y + z = 7

2x - y + 2z = 4

x + y - 2z = -1

Solution :

 1 2 1 2 -1 2 1 1 -2

 x y z

=

 7 4 -1

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

|A|

=

 1 2 1 2 -1 2 1 1 -2 |A|  =  2 [2-2] -2[-4-2]+1[2+1]

=  2(0) - 2(-6) + 1(3)

=  0+12+3

|A|  =  15 ≠ 0

Since A is a non singular matrix. A⁻¹ exists. x y z

= 1/15

 0 5 5 6 -3 0 3 1 -5

 7 4 -1

 x  =  15/15x  =  1 y  =  30/15y  =  2 z  =  30/15z  =  2

Example 3 :

Solve the following linear equation by inversion method

2x + y + z = 5

x + y + z = 4

x - y + 2z = 1

Solution :

 2 1 1 1 1 1 1 -1 2

 x y z

=

 5 4 1

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

|A|

=

 2 1 1 1 1 1 1 -1 2 =  2(2+1)+1(2-1)+1(-1-1)

=  2(3)-1-2

=  3

|A|  =  3  ≠  0

Since A is a non singular matrix. A⁻¹ exists x y z

= 1/3

 3 -3 0 -1 3 -1 -2 3 1

 5 4 1

 x  =  3/3x  =  1 y  =  6/3y  =  2 z  =  30/15z  =  2

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