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# CramerŌĆÖs Rule

i. Theorem (without proof) CramerŌĆÖs Rule, ii. Non Homogeneous linear equations upto three variables

CramerŌĆÖs Rule

Gabriel Cramer, a Swiss mathematician born in the city Geneva in 31 July 1704. He edited the works of the two elder BernoulliŌĆÖs, and wrote on the physical cause of the spheriodal shape of the planets and the motion of their apsides (1730), and on NewtonŌĆÖs treatment of cubic curves (1746).

In 1750 he published CramerŌĆÖs Rule, giving a general formula for the solution of certain linear system of n equations in n unknowns having a unique solution in terms of determinants. Advantages of CramerŌĆÖs rule is that we can find the value of x, y or z without knowing any of the other values of x, y or z. CramerŌĆÖs rule is applicable only when D ŌēĀ 0 ( is the determinant value of the coefficient matrix) for unique solution.

## Theorem (without proof) CramerŌĆÖs Rule:

Consider, If AX = B is a system of n linear equations in ŌĆśnŌĆÖ unknowns such that det (A) ŌēĀ 0, then the system has a unique solution.

This solution is, where Aj is the matrix obtained by replacing the entries in the jth column of A by the entries in the matrix ## 1. Non Homogeneous linear equations upto three variables.

(a) Consider the system of two linear equations with two unknowns. The solution of unknown by CramerŌĆÖs rule is (b) Consider the system of three linear equations with three unknowns a1

x + b1 y + c1 z = d1

a2 x + b2 y + c 2 z = d2

a3 x + b3 y + c 3 z = d3 Example 1.19

Solve the equations 2x + 3y = 7, 3x + 5y = 9 by CramerŌĆÖs rule.

Solution:

The equations are

2x + 3y = 7

3x + 5y = 9

Here Ōł┤ we can apply CramerŌĆÖs Rule Ōł┤ By CramerŌĆÖs rule Ōł┤    Solution is      x = 8, y = ŌłÆ3

Example 1.20

The following table represents the number of shares of two companies A and B during the month of January and February and it also gives the amount in rupees invested by Ravi during these two months for the purchase of shares of two companies. Find the the price per share of A and B purchased during both the months Solution:

Let the price of one share of A be x

Let the price of one share of B be y

Ōł┤ By given data, we get the following equations

10x + 5y = 125

9x + 12 y = 150 Ōł┤ By CramerŌĆÖs rule The price of the share A is Ōé╣10 and the price of the share B is Ōé╣5.

Example 1.21

The total cost of 11 pencils and 3 erasers is Ōé╣ 64 and the total cost of 8 pencils and 3 erasers is Ōé╣49. Find the cost of each pencil and each eraser by CramerŌĆÖs rule.

Solution:

Let ŌĆśxŌĆÖ be the cost of a pencil

Let ŌĆśyŌĆÖ be the cost of an eraser

Ōł┤ By given data, we get the following equations

11x + 3y =64

8x+3y=49 Ōł┤ The cost of a pencil is Ōé╣ 5 and the cost of an eraser is Ōé╣ 3.

Example 1.22

Solve by CramerŌĆÖs rule x + y + z = 4, 2x ŌłÆ y + 3z = 1, 3x + 2 y ŌłÆ z = 1

Solution: Ōł┤ We can apply CramerŌĆÖs Rule and the system is consistent and it has unique solution. The solution is     ( x , y, z) = ( ŌłÆ1,3, 2)

Example 1.23

The price of 3 Business Mathematics books, 2 Accountancy books and one Commerce book is Ōé╣840. The price of 2 Business Mathematics books, one Accountancy book and one Commerce book is Ōé╣570. The price of one Business Mathematics book, one Accountancy book and 2 Commerce books is Ōé╣630. Find the cost of each book by using CramerŌĆÖs rule.

Solution:

Let ŌĆśxŌĆÖ be the cost of a Business Mathematics book

Let ŌĆśyŌĆÖ be the cost of a Accountancy book.

Let ŌĆśzŌĆÖ be the cost of a Commerce book.

Ōł┤ 3x + 2 y + z = 840

2x + y + z = 570

x + y + 2z = 630 Ōł┤ The cost of a Business Mathematics book is Ōé╣120,

the cost of a Accountancy book is Ōé╣150 and

the cost of a Commerce book is Ōé╣180.

Example 1.24

An automobile company uses three types of Steel S1, S 2 and S3 for providing three different types of Cars C1, C2 and C3. Steel requirement R (in tonnes) for each type of car and total available steel of all the three types are summarized in the following table. Determine the number of Cars of each type which can be produced by CramerŌĆÖs rule.

Solution:

Let ŌĆśxŌĆÖ be the number of cars of type C1

Let ŌĆśyŌĆÖ be the number of cars of type C2

Let ŌĆśzŌĆÖ be the number of cars of type C3

3x + 2 y + 4z = 28

x + y + 2z = 13

2x + 2 y + z = 14 Ōł┤ The number of cars of each type which can be produced are 2, 3 and 4.

Tags : Theorem with Solved Example Problems , 12th Business Maths and Statistics : Chapter 1 : Applications of Matrices and Determinants
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12th Business Maths and Statistics : Chapter 1 : Applications of Matrices and Determinants : CramerŌĆÖs Rule | Theorem with Solved Example Problems