METHODS FOR SOLVING TWO EQUATIONS
METHODS FOR SOLVING THREE EQUATIONS
EXAMPLES
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APPLIED MATHEMATICS IN CHEMICAL ENGINEERING (HCE21 ) SEMINAR ON: CRAMER’S RULE DONE BY KISHAN KASUNDRA 4/21/2017 1 CHEMICAL ENGINEERING DEPARTMENT, DSCE
METHOD FOR SOLVING TWO EQUATIONS :- To find out variables, we need to have equations which are useful to solve the values of variables. First we assume two equations, a 1 x +b 1 y = c 1 a 2 x + b 2 y = c 2 Then we need to find determinant ‘D’ of coefficient matrix, for that construct 2x2 matrix Now, for determinant first we need to solve this matrix by cross multiplication, 4/21/2017 2 CHEMICAL ENGINEERING DEPARTMENT, DSCE
Now, replace the x-coefficient column with the constant column in coefficient matrix. Now, replace the y-coefficient column with the constant column in coefficient matrix. Now, according to Cramer’s rule, x= 𝑑𝑒𝑡𝐷𝑥/𝑑𝑒𝑡𝐷 & y = 𝑑𝑒𝑡𝐷𝑦/𝑑𝑒𝑡𝐷 4/21/2017 3 CHEMICAL ENGINEERING DEPARTMENT, DSCE
METHOD FOR SOLVING THREE EQUATIONS:- To find out variables, we need to have equations which are useful to solve the values of variables. First we assume two equations, a1x +b1y + c1z = d1 a2x + b2y +c2z = d2 a3x + b3y + c3z = d3 First, we need to find determinant ‘D’ of coefficient matrix, for that construct 3x3 matrix 4/21/2017 4 CHEMICAL ENGINEERING DEPARTMENT, DSCE
Now, for determinant first we need to solve this matrix, det D= a1(b2c3 – b3c2) – b1(a2c3 – c2a3) + c1 (a2b3 – b2a3 ) Now, replace the x-coefficient column with the constant column in coefficient matrix 4/21/2017 5 CHEMICAL ENGINEERING DEPARTMENT, DSCE
Now, replace the y-coefficient column with the constant column in coefficient matrix . det Dy = a1(d2c3 – d3c2) – d1(a2c3 – c2a3) + c1 (a2d3 – d2a3) 4/21/2017 6 CHEMICAL ENGINEERING DEPARTMENT, DSCE
Now, replace the z-coefficient column with the constant column in coefficient matrix . det Dz = a1(b2c3 – b3c2) – b1(a2c3 – c2a3) + c1 (a2b3 – b2a3) Now , according to Cramer’s rule, x= 𝑑𝑒𝑡𝐷𝑥/𝑑𝑒𝑡𝐷 y = 𝑑𝑒𝑡𝐷𝑦/𝑑𝑒𝑡𝐷 z = 𝑑𝑒𝑡𝐷z/𝑑𝑒𝑡𝐷 4/21/2017 7 CHEMICAL ENGINEERING DEPARTMENT, DSCE
EXAMPLES:- 1) Solve the system by using Cramer’s rule: 3x – 2y = 4 2x + y = -3 SOLUTION :- First find the determinants D, Dx , and Dy : D=( 3)(1) – (-2)(2) = 7 4/21/2017 8 CHEMICAL ENGINEERING DEPARTMENT, DSCE
Dx = (4)(1) – (-2)(-3) = - 2 Dy = (3 )(-3) – (4)(2) = - 17 By Cramer’s rule, we have x = 𝐷𝑥/𝐷 = − 2/7 y = 𝐷𝑦/𝐷 = − 17/7 The solution is ( x,y ) = (-2/7, -17/7). 4/21/2017 9 CHEMICAL ENGINEERING DEPARTMENT, DSCE