CRAMER'S RULE FOR A 2x2 LINEAR SYSTEM

Consider the following system of linear equation.

a1x + b1y = c1

a2x + b2y = c2

Write a determinant Δ with the coefficients of x and y as shown below.

To get the determinant Δx, replace the first column elements a1, a2 of Δ with c1, c2 respectively and keep the second column elements b1, bof Δ as it is.

To get the determinant Δy, replace the second column elements b1, b2 of Δ with c1, c2 respectively and keep the first column elements a1, aof Δ as it is.

By Cramer's Rule,

Example :

Solve the following system of linear equations using Cramer's rule.

x + 2y = 3

x + y = 2

Solution :

Find the value of Δ :

Δ = 1 - 2

Δ = -1

Find the value of Δ:

Δ= 3 - 4

Δ= -1

Find the value of Δ:

Δ= 2 - 3

Δ= -1

By Cramer's Rule,

x = 1,  y = 1

Note :

Cramer’s rule is applicable when Δ ≠ 0.

If Δ = 0, then the given system may be consistent or inconsistent.

Case 1 :

If Δ = 0 and Δx = 0, Δy = 0 and atleast one of the coefficients a11, a12, a21, a22 is non-zero, then the system is consistent and has infinitely many solutions.

Case 2 :

If Δ = 0 and atleast one of the values Δx, Δy is non-zero, then the system is inconsistent i.e. it has no solution.

Solve the following systems using Cramer's Rule.

Problem 1 :

3x + 2y = 5

x + 3y = 4

Solution :

Find the value of Δ :

Δ = 9 - 2

Δ = 7

Find the value of Δ:

 Δ= 15 - 8

 Δ= 7

Find the value of Δ:

Δ= 12 - 5

Δ= 7

By Cramer's Rule,

x = 1,  y = 1

Problem 2 :

x + 2y = 3

2x + 4y = 8

Solution :

Find the value of Δ :

Δ = 4 - 4

Δ = 0

Find the value of Δ:

Δ= 12 - 16

Δ= -4

Since Δ = 0 and Δ≠ 0, the system is consistent and it has no solution. 

Problem 3 :

x + 2y = 3

2x + 4y = 6

Solution :

Find the value of Δ :

Δ = 4 - 4

Δ = 0

Find the value of Δ:

Δ= 12 - 12

Δ= 0

Find the value of Δ:

Δ= 6 - 6

Δ= 0

Since ∆ = 0, ∆x = 0, = 0 and ∆ has atleast one non-zero element, the system is consistent and it has infinitely many solutions.

Problem 4 :

2x + 4y = 6

6x + 12y = 24

Solution :

Find the value of Δ :

Δ = 24 - 24

Δ = 0

Find the value of Δ:

Δ= 72 - 96

Δ= -24

Since Δ = 0 and Δ≠ 0, the system is consistent and it has no solution.

Problem 5 :

2x + y = 3

6x + 3y = 9

Solution :

Find the value of Δ :

Δ = 6 - 6

Δ = 0

Find the value of Δ:

Δx = 9 - 9

Δx = 0

Find the value of Δ:

Δ= 18 - 18

Δ= 0

Since ∆ = 0, ∆x = 0, = 0 and ∆ has atleast one non-zero element, the system is consistent and it has infinitely many solutions.

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