CRAMER'S RULE OF SOLVING SIMULTANEOUS EQUATIONS

In this section, you will learn how to solve system of simultaneous equations using Cramer's rule. 

Let us consider the following system of three equations with three unknowns x, y and z. 

a₁₁x + a₁₂y + a₁₃z  =  b₁

a₂₁x + a₂₂y + a₂₃z  =  b₂

a₃₁x + a₃₂y + a₃₃z  =  b₁

Now, we can write the the following determinants using the above equations. 

Then, Cramer’s rule to find the values of x, y and z :

x  =  Δ₁/Δ

y  =  Δ₂/Δ

z  =  Δ₃/Δ

If Δ  =  0, the system is inconsistent and it has solution. 

Solving System of Simultaneous Equations Using Cramer's Rule - Examples

Example 1 :

Solve the following system of linear equations using Cramer’s rule:

5x − 2y + 16  =  0

x + 3y − 7  =  0

Solution :

5x − 2y  =  -16 ------(1) 

x + 3y  =  7 ------(2) 

By Cramer's rule, 

x  =  Δ₁/Δ  =  -34/17  =  -2

y  =  Δ₂/Δ  =  51/17  =  3

So, the values of x and y are 2 and 3 respectively.

Example 2 :

Solve the following system of linear equations using Cramer’s rule :

(3/x) + 2y  =  12

(2/x) + 3y  =  13

Solution :

Let 1/x  =  x_1.

Then, 

3x₁ + 2y  =  12

2x₁ + 3y  =  13

By Cramer's rule, 

x₁  =  Δ₁/Δ  =  10/5  =  2

1/x  =  2 -----> x  =  1/2

y  =  Δ₂/Δ  =  15/5  =  3

So, the values of x and y are 1/2 and 3 respectively.

Example 3 :

Solve the following system of linear equations using Cramer’s rule :

3x + 3y − z  =  11

2x − y + 2z  =  9

4x + 3y + 2z  =  25

Solution :

Δ  =  3(-2-6) - 3(4-8) - 1(6+4)

  =  3(-8) - 3(-4) - 1(10)

  =  -24 + 12 - 10

  =  -34 + 12

  =  -22

Δ₁ =  11(-2-6) - 3(18-50) - 1(27+25)

  = 11(-8) - 3(-32) - 1(52)

=  -88 + 96 - 52

=  -140 + 96

Δ₁  =  -44

Δ₂ =  3(18 - 50) - 11(4 - 8) - 1(50 - 36)

  =  3(-32) - 11(-4) - 1(14)

  =  -96 + 44 - 14

  =  -110 + 44

Δ₂  =  -66

Δ₃ =  3(-25 - 27) - 3(50 - 36) + 11(6 + 4)

  =  3(-52) - 3(14) + 11(10)

  =  -156 - 42 + 110

  =  -198 + 110

Δ₃  =  -88

By Cramer's rule, 

x  =  Δ₁/Δ  =  -44/(-22)  =  2

y  =  Δ₂/Δ  =  -66/(-22)  =  3

z  =  Δ₃/Δ  =  -88/(-22)  =  4

So, the values of x, y and z are 2, 3 and 4 respectively.

Example 4 :

Solve the following system of linear equations using Cramer’s rule :

(3/x) - (4/y) - (2/z) - 1  =  0

(1/x) + (2/y) + (1/z) - 2  =  0

(2/x) - (5/y) - (4/z) + 1  =  0

Solution :

Let 1/x  =  a, 1/y  =  b and 1/z  =  c

3a - 4b + 2c  =  1 -----(1)

a + 2b + c  =  2 -----(2)

2a - 5b - 4c  =  -1 -----(3)

Δ  =  3(-8+5) + 4(-4-2) - 2(-5-4)

  =  3(-3) + 4(-6) - 2(-9)

  =  -9 - 24 + 18

Δ  =  -15

Δ₁ =  1(-8+5) + 4(-8+1) -2(-10+2)

  =  1(-3) + 4(-7) - 2(-8)

  =  -3 - 28 + 16

Δ₁  =  -15

Δ₂  =  3(-8+1) - 1(-4-2) - 2(-1-4)

  =  3(-7) - 1(-6) - 2(-5)

  =  -21 + 6 + 10

  =  -21 + 16

 Δ₂  =  -5

Δ₃ =  3(-2+10) + 4(-1-4) + 1(-5-4)

  =  3(8) + 4(-5) + 1(-9)

  =  24 - 20 - 9

  =  -5

Δ₃  =  -5

By Cramer's rule, 

a  =  Δ₁/Δ  =  -15/(-15)  =  1

b  =  Δ₂/Δ  =  -5/(-15)  =  1/3

c  =  Δ₃/Δ  =  -5/(-15)  =  1/3

Then, 

x  =  1/a  =  1/1  =  1

y  =  1 / (1/3)  =  3

z  =  1 / (1/3)  =  3

So, the values of x, y and z are 1, 3 and 3 respectively. 

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