1. 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. 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. 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 6 3 = (2)(3) –(6)(-1) =6+6 =12 7 -14 -2 4 = (7)(4) –(-2)(-14) =28-28 =0 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 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. 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= 2 1 5 1 4 1 7 1 1 1 y= 2 1 5 1 2 4 5 7 = = (4)(1) –(7)(1) (2)(1) –(5)(1) (2)(7) –(5)(4) = = 14 - 20 4 - 7 2 - 5 = -6 -3 =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. 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= 4 2 5 1 10 2 17 1 y= 4 2 5 1 4 10 5 17 = = (10)(1) –(17)(2) (4)(1) –(5)(2) (4)(17) –(5)(10) = = 68 - 50 10 - 34 4 - 10 = 18 -6 = -3 The system is consistent and independent, with a unique solution at (4,-3) = -24 -6 = 4 4x + 2y = 10 5x + y = 17 1 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 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 6 4 3 2 10 4 5 2 = (10)(2) –(5)(4) (6)(2) –(3)(4) = 20 - 20 12 - 12 = 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. 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= 4.6 2.4 8.2 4.2 2.4 28.6 y= 4.6 2.4 8.2 4.6 4.2 8.2 28.6 = = (4.2)(-1) –(28.6)(2.4) (4.6)(-1) –(8.2)(2.4) (4.6)(28.6) –(8.2)(4.2) = = 131.56-34.44 -4.2 - 68.64 -4.6 - 19.68 = 97.12 -24.28 = -4 The system is consistent and independent, with a unique solution at (3,-4) = -72.84 -24.28 = 3 4.6x + 2.4y = 4.2 8.2x - y = 28.6 1 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. 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 = 4 -2 3 3 -6x - 4y + 6z = 4 12x +9y -6z = 12 6x + 5y = 16 4x + 3y -2z = 4 5x + 4y -6z = -5 -3 -12x - 9y + 6z = -12 5x + 4y - 6z = -5 -7x -5y = -17 6x + 5y = 16 -7x -5y = -17 - x = -1 x=1 6x + 5y = 16 6( )+ 5y = 16 1 6 + 5y = 16 -6 5y = 10 5 y = 2 3( ) + 2( ) -3z = -2 1 2 3 + 4 – 3z = -2 7 – 3z = -2 -7 -3z = -9 -3 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. 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 = 12 -3 2 2 -6x - 9y + 6z = -15 10x +6y -6z = 24 4x - 3y = 9 5x + 3y -3z = 12 4x + 2y -5z = 4 -5 3 3 -25x - 15y + 15z = -60 12x + 6y -15z = 12 -13x - 9y = -48 4x - 3y = 9 -13x -9y = -48 -3 -12x + 9y = -27 -13x - 9y = -48 -25x = -75 -25 x=3 4x - 3y = 9 4( )- 3y = 9 3 12 - 3y = 9 -12 -3y = -3 -3 y = 1 2( ) + 3( ) -2z = 5 3 1 6 + 3 – 2z = 5 9 – 2z = 5 -9 -2z = -4 -2 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