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

2 Standard 2: 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. ALGEBRA II STANDARDS THIS LESSON AIMS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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

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

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

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

7 Standards 2, 25 Solve the following system of equations using Cramer’s Rule: ax + by = e cx + dy = f x= a b c d e b f d = (10)(2) –(5)(4) (6)(2) –(3)(4) = = 0 0 6x + 4y = 10 3x + 2y = 5 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

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

9 Standard 5 SYSTEMS OF EQUATIONS IN THREE VARIABLES 3x + 2y -3z = -2 The system has unique solution at (1,2,3) 4x + 3y -2z = 4 5x + 4y -6z = -5 Solve: 3x + 2y -3z = -2 4x + 3y -2z = x - 4y + 6z = 4 12x +9y -6z = 12 6x + 5y = 16 4x + 3y -2z = 4 5x + 4y -6z = x - 9y + 6z = -12 5x + 4y - 6z = -5 -7x -5y = -17 6x + 5y = 16 -7x -5y = x = -1 x=1 6x + 5y = 16 6( )+ 5y = y = y = 10 5 y = 2 3( ) + 2( ) -3z = – 3z = -2 7 – 3z = z = z = 3 3x + 2y -3z = -2 Eliminating one variable, this case z: 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: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10 Standard 5 SYSTEMS OF EQUATIONS IN THREE VARIABLES 2x + 3y -2z = 5 The system has unique solution at (3,1,2) 5x + 3y -3z = 12 4x + 2y -5z = 4 Solve: 2x + 3y -2z = 5 5x + 3y -3z = x - 9y + 6z = x +6y -6z = 24 4x - 3y = 9 5x + 3y -3z = 12 4x + 2y -5z = x - 15y + 15z = x + 6y -15z = x - 9y = -48 4x - 3y = 9 -13x -9y = x + 9y = x - 9y = x = x=3 4x - 3y = 9 4( )- 3y = y = y = y = 1 2( ) + 3( ) -2z = – 2z = 5 9 – 2z = z = z = 2 2x + 3y -2z = 5 Eliminating one variable, this case z: 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: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved