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 Δor 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 Δ:

 Δ= 5(2 + 1) - 1(8 - 1) + 1(-4 - 1)

 Δ= 5(3) - 1(7) + 1(-5)

 Δ= 15 - 7 - 5

 Δ= 3

Find the value of Δ:

Δ= 2(8 - 1) - 5(2 - 1) + 1(1 - 4)

Δ= 2(7) - 5(1) + 1(-3)

Δ= 14 - 5 - 3

Δ= 6

Find the value of Δ:

Δ= 2(1 + 4) - 1(1 - 4) + 5(-1 - 1)

Δ= 2(5) - 1(-3) + 5(-2)

Δ= 10 + 3 - 10

Δ= 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:

 Δ= 0

Find the value of Δ:

Δ= 0

Find the value of Δ:

Δ= 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 Δ:

 Δ= 0

Find the value of Δ:

Δ= 0

Find the value of Δ:

Δ= 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 Δ:

 Δ≠ 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 Δ:

Δ= 0

Find the value of Δ:

Δ= 0

Find the value of Δ:

Δ= 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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