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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 SHOW1
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ALGEBRA II STANDARDS THIS LESSON AIMS:
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.
2
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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
= (2)(3) –(6)(-1)=6+6 =12
7
-14
-2
4
= (7)(4) –(-2)(-14) =28-28 =0
3
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Standards 2, 25
CRAMER’S RULE
ax + by = e
The solution to the system
e
b
f
d
a
b
c
d
is (x,y)
cx + dy = f
x=
where
a
b
and
=0
c
d
a e
c
f
a
b
c
d
y=
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4
Solve the following system of equations using Cramer’s Rule:
ax + by = e
cx + dy = f
2x +1y = 4
5x + 1y = 7
e
b
4 1
f
d
7
x=
y=
(4)(1) –(7)(1)
1
x=
=
a
b
2 1
c
d
5
(2)(1) –(5)(1)
=
4 -7
=
-3
=1
-3
=
-6
=2
-3
2 - 5
1
a e
2 4
c
f
5
a
b
2 1
c
d
5
y=
Standards 2, 25
7
(2)(7) –(5)(4)
=
(2)(1) –(5)(1)
=
14 - 20
2 - 5
1
The system is consistent and independent,
with a unique solution at (1,2)
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5
Solve the following system of equations using Cramer’s Rule:
ax + by = e
cx + dy = f
4x + 2y = 10
5x + 1y = 17
e
b
10 2
f
d
17 1
x=
y=
(10)(1) –(17)(2)
x=
=
a
b
4
2
c
d
5
1
a e
4
10
c
f
5
17
a
b
4
2
c
d
5
1
y=
Standards 2, 25
(4)(1) –(5)(2)
=
(4)(17) –(5)(10)
=
(4)(1) –(5)(2)
=
10 - 34
=
-24
=4
-6
=
18
= -3
-6
4 - 10
68 - 50
4 - 10
The system is consistent and independent,
with a unique solution at (4,-3)
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6
Solve the following system of equations using Cramer’s Rule:
ax + by = e
cx + dy = f
6x + 4y = 10
3x + 2y = 5
e
b
10 4
f
d
5 2
x=
Standards 2, 25
x=
(10)(2) –(5)(4)
=
a
b
6
4
c
d
3
2
(6)(2) –(3)(4)
=
20 - 20
0
= 0
12 - 12
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.
7
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Solve the following system of equations
Standards 2, 25
ax + by = e
cx + dy = f
4.6x + 2.4y = 4.2
8.2x -1y = 28.6
This is a typical case when it is useful to use Cramer’s rule, because
substitution or elimination are both difficult to apply.
e
b
4.2 2.4
f
d
28.6 -1
a
b
4.6
(4.2)(-1) –(28.6)(2.4) -4.2 - 68.64 -72.84
=3
=
=
=
-24.28
(4.6)(-1) –(8.2)(2.4) -4.6 - 19.68
2.4
c
d
8.2
-1
x=
y=
x=
a e
4.6 4.2
c
f
8.2
a
b
4.6
c
d
8.2
y=
28.6 (4.6)(28.6) –(8.2)(4.2) 131.56-34.44 97.12
= -4
=
=
=
-24.28
-4.6 -19.68
2.4 (4.6)(-1) –(8.2)(2.4)
-1
The system is consistent and independent,
with a unique solution at (3,-4)
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8
Standard 5
SYSTEMS OF EQUATIONS IN THREE VARIABLES
Eliminating one variable, this case z:
Solve:
3x + 2y -3z = -2
4x + 3y -2z = 4
5x + 4y -6z = -5
-2 3x + 2y -3z = -2 -2
3 4x + 3y -2z = 4 3
-3 4x + 3y -2z = 4
5x + 4y -6z = -5
-3
Using the equations with
two variables to eliminate Using one of the two
other variable, this case y: variable equations to
substitute x and get the
6x + 5y = 16
other variable, this case y:
-7x -5y = -17
6x + 5y = 16
- x = -1
6( 1 )+ 5y = 16
-1 -1
6 + 5y = 16
-6
-6
x=1
5y = 10
5
5
y=2
-6x - 4y + 6z = 4
12x +9y -6z = 12
6x + 5y = 16
-12x - 9y + 6z = -12
5x + 4y - 6z = -5
-7x -5y = -17
Using one of the three variable
equations to substitute x and y to
get z:
3x + 2y -3z = -2
3( 1 ) + 2( 2 ) -3z = -2
The system has unique solution at (1,2,3)
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3 + 4 – 3z = -2
7 – 3z = -2
-7
-7
-3z = -9
-3 -3
z=3
9
Standard 5
SYSTEMS OF EQUATIONS IN THREE VARIABLES
Eliminating one variable, this case z:
Solve:
2x + 3y -2z = 5
5x + 3y -3z = 12
4x + 2y -5z = 4
-3 2x + 3y -2z = 5 -3
2 5x + 3y -3z = 12 2
-6x - 9y + 6z = -15
10x +6y -6z = 24
4x - 3y = 9
-5 5x + 3y -3z = 12 -5
3 4x + 2y -5z = 4 3
-25x - 15y + 15z = -60
12x + 6y -15z = 12
-13x - 9y = -48
Using the equations with
two variables to eliminate Using one of the two
other variable, this case y: variable equations to
substitute x and get the
-3 4x - 3y = 9 -3
other variable, this case y:
-13x -9y = -48
4x - 3y = 9
-12x + 9y = -27
4( 3 )- 3y = 9
-13x - 9y = -48
12 - 3y = 9
-25x = -75
-12
-12
-25 -25
-3y = -3
x=3
-3 -3
y=1
Using one of the three variable
equations to substitute x and y to
get z:
2x + 3y -2z = 5
2( 3 ) + 3( 1 ) -2z = 5
The system has unique solution at (3,1,2)
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6 + 3 – 2z = 5
9 – 2z = 5
-9
-9
-2z = -4
-2 -2
z=2
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