Consider the following system of linear equation.
a_{1}x + b_{1}y + c_{1}z = d_{1}
a_{2}x + b_{2}y + c_{2}z = d_{2}
a_{3}x + b_{3}y + c_{3}z = d_{3}
Write a determinant Δ with the coefficients of x, y and z as shown below.
To get the determinant Δ_{x}, replace the first column elements a_{1}, a_{2}, a_{3} of Δ with d_{1}, d_{2}, d_{3} respectively and keep the second and third columns as they are.
To get the determinant Δ_{y} replace the second column elements b_{1}, b_{2}, b_{3} of Δ with d_{1}, d_{2}, d_{3} respectively and keep the first and third columns as they are.
To get the determinant Δ_{x}, replace the third column elements c_{1}, c_{2}, c_{3} of Δ with d_{1}, d_{2}, d_{3} respectively and keep the first and second columns as they are.
By Cramer's Rule,
Note :
Cramer’s rule is applicable when Δ ≠ 0.
When Δ ≠ 0, the system has unique solution or only one solution.
If Δ = 0, then the given system may be consistent or inconsistent.
When Δ = 0, we have the following four possibilities.
Case (i) :
If Δ = 0 and Δ_{x }or Δ_{y} or Δ_{z} ≠ 0, then the system is consistent and has NO solution.
Case (ii) :
If Δ = 0 and Δ_{x} = Δ_{y} = Δ_{z} = 0 and atleast one of the 2 × 2 minor of Δ is non zero, then the system is consistent and has infinitely many solution.
Case (iii) :
If Δ = 0 and Δ_{x} = Δ_{y} = Δ_{z} = 0 and all their (2 × 2) minors are zero but atleast one of the elements of Δ is non zero, then the system is consistent and has infinitely many solution.
Case (iv) :
If Δ = 0, Δ_{x} = Δ_{y} = Δ_{z} = 0, all 2 × 2 minors of Δ = 0 and atleast one 2 × 2 minor of Δ_{x} or Δ_{y} or Δ_{z} is non zero then the system is inconsistent and has NO solution.
Solve the following systems using Cramer's Rule.
Example 1 :
2x + y + z = 5
x + y + z = 4
x - y + 2z = 1
Solution :
Find the value of Δ :
Δ = 2(2 + 1) - 1(2 - 1) + 1(-1 - 1)
Δ = 2(3) - 1(1) + 1(-2)
Δ = 6 - 1 - 2
Δ = 3 ≠ 0
Since Δ ≠ 0, the system has only one solution.
Find the value of Δ_{x }:
Δ_{x }= 5(2 + 1) - 1(8 - 1) + 1(-4 - 1)
Δ_{x }= 5(3) - 1(7) + 1(-5)
Δ_{x }= 15 - 7 - 5
Δ_{x }= 3
Find the value of Δ_{y }:
Δ_{y }= 2(8 - 1) - 5(2 - 1) + 1(1 - 4)
Δ_{y }= 2(7) - 5(1) + 1(-3)
Δ_{y }= 14 - 5 - 3
Δ_{y }= 6
Find the value of Δ_{z }:
Δ_{z }= 2(1 + 4) - 1(1 - 4) + 5(-1 - 1)
Δ_{z }= 2(5) - 1(-3) + 5(-2)
Δ_{z }= 10 + 3 - 10
Δ_{z }= 3
By Cramer's Rule,
x = 1, y = 2, z = 1
Therefore,
(x, y, z) = (1, 2, 1)
Example 2 :
x + 2y + 3z = 6
x + y + z = 3
2x + 3y + 4z = 9
Solution :
Find the value of Δ :
Δ = 0
Find the value of Δ_{x }:
Δ_{x }= 0
Find the value of Δ_{y }:
Δ_{y }= 0
Find the value of Δ_{z }:
Δ_{z }= 0
Δ = 0 and Δ_{x} = Δ_{y} = Δ_{z} = 0, but atleast one of the 2 × 2 minors of Δ is non-zero.
= 1 - 2
= -1 ≠ 0
Therefore, the system is consistent (by case (ii)) and has infinitely many solutions.
Example 3 :
x + 2y + 3z = 6
2x + 4y + 6z = 12
3x + 6y + 9z = 18
Solution :
Find the value of Δ :
Δ = 0
Find the value of Δ_{x }:
Δ_{x }= 0
Find the value of Δ_{y }:
Δ_{y }= 0
Find the value of Δ_{z }:
Δ_{z }= 0
Δ = 0 and Δ_{x} = Δ_{y} = Δ_{z} = 0,
Also all the 2 × 2 minors of Δ, Δ_{x}, Δ_{y} and Δ_{z} are zero, but atleast one of the elements of Δ is non-zero.
Therefore, the system is consistent (by case (iii)) and has infinitely many solutions.
Example 4 :
x + 2y + 3z = 6
x + y + z = 3
2x + 3y + 4z = 10
Solution :
Find the value of Δ :
Δ = 0
Find the value of Δ_{x }:
Δ_{x }≠ 0
Δ = 0, but Δ_{x} ≠ 0 (atleast one of the values of Δ_{x}, Δ_{y}, and Δ_{z} is non-zero).
Therefore, the system is inconsistent (by case (i)) and has NO solution.
Example 5 :
x + 2y + 3z = 6
2x + 4y + 6z = 12
3x + 6y + 9z = 24
Solution :
Find the value of Δ :
Δ = 0
Find the value of Δ_{x }:
Δ_{x }= 0
Find the value of Δ_{y }:
Δ_{y }= 0
Find the value of Δ_{z }:
Δ_{z }= 0
Δ = 0 and Δ_{x} = Δ_{y} = Δ_{z} = 0,
All the 2 × 2 minors of Δ are zero, but we see that atleast one of the 2 × 2 minors of Δ_{x} or Δ_{y} or Δ_{z} is non-zero.
Therefore, the system is inconsistent (by case (i)) and has No solution.
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