SOLVING EQUATIONS USING DETERMINANTS Standards 2, 25 SOLVING EQUATIONS USING DETERMINANTS SECOND ORDER DETERMINANT CRAMER’S RULE PROBLEM 1 PROBLEM 2 PROBLEM 3 PROBLEM 4 SOLVING SYSTEMS OF EQUATIONS IN THREE VARIABLES PROBLEM 5 PROBLEM 6 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

ALGEBRA II STANDARDS THIS LESSON AIMS: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. Estándar 2: Los estudiantes resuelven sistemas de ecuaciones lineares y desigualdades (en 2 o tres variables) por substitución, con gráficas o con matrices. Standard 25: Students use properties from number systems to justify steps in combining and simplifying functions. Estándar 25: Los estudiantes usan propiedades de sistemas numéricos para justificar pasos en combinar y simplificar funciones. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SECOND ORDER DETERMINANT Standards 2, 25 SECOND ORDER DETERMINANT a b c d rows columns VALUE OF A SECOND ORDER DETERMINANT a b c d = ad - cb Find the value of the following determinants 2 -1 6 3 7 -14 -2 4 = (2)(3) –(6)(-1) =6+6 =12 = (7)(4) –(-2)(-14) =28-28 =0 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

The solution to the system is (x,y) cx + dy = f Standards 2, 25 CRAMER’S RULE ax + by = e The solution to the system is (x,y) cx + dy = f e b f d x= a b c d where a b c d = 0 and a e c f y= a b c d PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Solve the following system of equations using Cramer’s Rule: Standards 2, 25 2x + y = 4 ax + by = e 1 5x + y = 7 1 cx + dy = f x= a b c d e f 4 1 7 (4)(1) –(7)(1) 4 - 7 = -3 x= = = =1 (2)(1) –(5)(1) 2 - 5 2 1 5 y= a b c d e f 2 4 5 7 (2)(7) –(5)(4) 14 - 20 = -6 -3 y= = = =2 (2)(1) –(5)(1) 2 - 5 2 1 5 The system is consistent and independent, with a unique solution at (1,2) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Solve the following system of equations using Cramer’s Rule: Standards 2, 25 4x + 2y = 10 ax + by = e 5x + y = 17 1 cx + dy = f x= a b c d e f 10 2 17 1 (10)(1) –(17)(2) 10 - 34 = -24 -6 x= = 4 = = 4 2 5 1 (4)(1) –(5)(2) 4 - 10 y= a b c d e f 4 10 5 17 (4)(17) –(5)(10) 68 - 50 = 18 -6 y= = = = -3 (4)(1) –(5)(2) 4 - 10 4 2 5 1 The system is consistent and independent, with a unique solution at (4,-3) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Solve the following system of equations using Cramer’s Rule: Standards 2, 25 6x + 4y = 10 ax + by = e 3x + 2y = 5 cx + dy = f x= a b c d e f 10 4 5 2 (10)(2) –(5)(4) 20 - 20 = x= = = (6)(2) –(3)(4) 12 - 12 6 4 3 2 Since, the determinant from the denominator is zero, and division by zero is not defined: THIS SYSTEM DOES NOT HAVE A UNIQUE SOLUTION and Cramer’s Rule can’t be used. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Solve the following system of equations Standards 2, 25 4.6x + 2.4y = 4.2 ax + by = e 8.2x - y = 28.6 1 cx + dy = f This is a typical case when it is useful to use Cramer’s rule, because substitution or elimination are both difficult to apply. x= a b c d e f 4.2 2.4 28.6 -1 (4.2)(-1) –(28.6)(2.4) -4.2 - 68.64 = -72.84 -24.28 x= = = = 3 (4.6)(-1) –(8.2)(2.4) -4.6 - 19.68 4.6 2.4 8.2 -1 4.6 4.2 8.2 28.6 y= a b c d e f (4.6)(28.6) –(8.2)(4.2) 131.56-34.44 = 97.12 -24.28 y= = = = -4 (4.6)(-1) –(8.2)(2.4) -4.6 -19.68 4.6 2.4 8.2 -1 The system is consistent and independent, with a unique solution at (3,-4) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SYSTEMS OF EQUATIONS IN THREE VARIABLES Standard 5 SYSTEMS OF EQUATIONS IN THREE VARIABLES Eliminating one variable, this case z: Solve: -2 3x + 2y -3z = -2 -6x - 4y + 6z = 4 12x +9y -6z = 12 3x + 2y -3z = -2 3 4x + 3y -2z = 4 6x + 5y = 16 4x + 3y -2z = 4 5x + 4y -6z = -5 -3 4x + 3y -2z = 4 -12x - 9y + 6z = -12 5x + 4y - 6z = -5 5x + 4y -6z = -5 -7x -5y = -17 Using the equations with two variables to eliminate other variable, this case y: Using one of the two variable equations to substitute x and get the other variable, this case y: Using one of the three variable equations to substitute x and y to get z: 6x + 5y = 16 -7x -5y = -17 3x + 2y -3z = -2 6x + 5y = 16 3( ) + 2( ) -3z = -2 1 2 - x = -1 6( )+ 5y = 16 1 3 + 4 – 3z = -2 -1 -1 6 + 5y = 16 7 – 3z = -2 x=1 -6 -6 -7 -7 5y = 10 -3z = -9 5 5 -3 -3 y = 2 z = 3 The system has unique solution at (1,2,3) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SYSTEMS OF EQUATIONS IN THREE VARIABLES Standard 5 SYSTEMS OF EQUATIONS IN THREE VARIABLES Eliminating one variable, this case z: Solve: -3 2x + 3y -2z = 5 -6x - 9y + 6z = -15 2 10x +6y -6z = 24 2x + 3y -2z = 5 5x + 3y -3z = 12 4x - 3y = 9 5x + 3y -3z = 12 4x + 2y -5z = 4 -5 5x + 3y -3z = 12 -25x - 15y + 15z = -60 3 12x + 6y -15z = 12 4x + 2y -5z = 4 -13x - 9y = -48 Using the equations with two variables to eliminate other variable, this case y: Using one of the two variable equations to substitute x and get the other variable, this case y: Using one of the three variable equations to substitute x and y to get z: -3 4x - 3y = 9 -13x -9y = -48 2x + 3y -2z = 5 4x - 3y = 9 2( ) + 3( ) -2z = 5 3 1 -12x + 9y = -27 4( )- 3y = 9 3 -13x - 9y = -48 6 + 3 – 2z = 5 12 - 3y = 9 -25x = -75 9 – 2z = 5 -12 -12 -25 -25 -9 -9 -3y = -3 x=3 -2z = -4 -3 -3 -2 -2 y = 1 z = 2 The system has unique solution at (3,1,2) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved