CRAMER'S RULE FOR A 3x3 LINEAR SYSTEM
Consider the following system of linear equation.
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Write a determinant Δ with the coefficients of x, y and z as shown below.
To get the determinant Δx, replace the first column elements a1, a2, a3 of Δ with d1, d2, d3 respectively and keep the second and third columns as they are.
To get the determinant Δy replace the second column elements b1, b2, b3 of Δ with d1, d2, d3 respectively and keep the first and third columns as they are.
To get the determinant Δx, replace the third column elements c1, c2, c3 of Δ with d1, d2, d3 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 Δx or Δy or Δz ≠ 0, then the system is consistent and has NO solution.
Case (ii) :
If Δ = 0 and Δx = Δy = Δz = 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 Δx = Δy = Δz = 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, Δx = Δy = Δz = 0, all 2 × 2 minors of Δ = 0 and atleast one 2 × 2 minor of Δx or Δy or Δz 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 Δx :
Δx = 5(2 + 1) - 1(8 - 1) + 1(-4 - 1)
Δx = 5(3) - 1(7) + 1(-5)
Δx = 15 - 7 - 5
Δx = 3
Find the value of Δy :
Δy = 2(8 - 1) - 5(2 - 1) + 1(1 - 4)
Δy = 2(7) - 5(1) + 1(-3)
Δy = 14 - 5 - 3
Δy = 6
Find the value of Δz :
Δz = 2(1 + 4) - 1(1 - 4) + 5(-1 - 1)
Δz = 2(5) - 1(-3) + 5(-2)
Δz = 10 + 3 - 10
Δz = 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 Δx :
Δx = 0
Find the value of Δy :
Δy = 0
Find the value of Δz :
Δz = 0
Δ = 0 and Δx = Δy = Δz = 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 Δx :
Δx = 0
Find the value of Δy :
Δy = 0
Find the value of Δz :
Δz = 0
Δ = 0 and Δx = Δy = Δz = 0,
Also all the 2 × 2 minors of Δ, Δx, Δy and Δz 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 Δx :
Δx ≠ 0
Δ = 0, but Δx ≠ 0 (atleast one of the values of Δx, Δy, and Δz 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 Δx :
Δx = 0
Find the value of Δy :
Δy = 0
Find the value of Δz :
Δz = 0
Δ = 0 and Δx = Δy = Δz = 0,
All the 2 × 2 minors of Δ are zero, but we see that atleast one of the 2 × 2 minors of Δx or Δy or Δz is non-zero.
Therefore, the system is inconsistent (by case (i)) and has No solution.
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