Download STUDY GUIDE FOR CHAPTER 6

STUDY GUIDE FOR CHAPTER 6
6.1. Systems of Linear Equations


In this section student should be bale to solve linear equations in two and three variables using:
 Addition – elimination method
 Substitution
 Using graphing calculator
Student should be able to recognize consistent, inconsistent and dependent system
Practice: Q. 23 on page 436 – use the elimination method
x y
 4
2 3
Multiply first eq. by (- 3)
3x 3 y

 15
2
2
3x
 y  12
2
Add two eq’s.
3x 3 y

 15
2
2

1
y  3 y  6
2
x 6
 4 x4
2 3
Practice – Use substitution and elimination method.
2 x  3 y  10
7x  2 y  3
1.
2.
 3x  y  18
6x  y  3
4.
3 x  2 y  6
4 x  y  14
5.
3.
3x  7 y  10
18 x  42 y  50
5 x  2 y  11
10 x  4 y  22
Practice: Q. 32 on page 436 – [System is Inconsistent]
3x  2 y  5
6x  4 y  8
Practice: Q. 33 on page 436 – [System is Dependent]
4x  y  9
 8 x  2 y  18
Practice: Q. 62 on page 438
3
p q
3
3
2
q  81  q

2
4
3
p  81  q
4
1
2 q  81
q  36


4

p
3
3
q  q  81 
2
4
3
3
q  36  54
2
4
1
Solving linear equations in three variables
Practice: Q. 40 on page 436 – Use substitution method.
4 x  y  3 z  2
3x  5 y  z  15
{(-1, 4, 2)}
Ordered triple
 2 x  y  4 z  14
3x  2 y  z  1
x  y  2 z  3
Use graphing calculator to solve the system.
2 x  3 y  z  8
Practice: Q. 55 on page 437
(2,3)
y  ax 2  bx  c
(-1,0) (-2,2)
Solution:
(2,3)
3  a  22  b  2  c
3  4 a  2b  c
(-1,0) 0  a  b  c
Use graphing calculator to solve the system
(-2,2) 2  4a  2b  c
Practice: Q. 56 on page 437
Practice: Q. 69 on page 439
6.2. Matrix Solution of Linear Systems

In this section student should be able to solve systems of linear equations in two and three variables using
Gauss-Jordan method.
Example:
1 x  2 y  5
2 x  1  y  2
Form the augmented matrix from variable coefficients:
1
2
5
2
1 2
Gauss-Jordan method requires writing the matrix in the following form:
1 0 a
Then a, and b are the solution of the system
0 1 b
Or for the system of three equations:
1 0 0 a
0 1 0 b
0 0 1 c
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Matrix Row Transformation:
1. Any two rows can be interchanged.
2. The elements of nay row may be multiplied by a nonzero real number.
3. Any row may be changed by adding to its elements a multiple of elements of another row.
1x  2 y  5
Practice: Solve the system of equations by performing raw transformation:
2 x  y  2
1 2 5
2 1 2
1 2
5
0  3  22
Change R2 with following (-2R1+R2)
1 2 5
0 1 4
1 0 3
Change R1 with following [-2R2+R1]
0 1 4
Practice:
3x  4 y  1
2 x  3 y  10
5 x  2 y  19
2x  2 y  5
Change R2 with following [R2/(-3)]
 x  3
 y4
4 x  2 y  3z  4
3x  5 y  z  7
5x  y  4 z  7
6.3. DETERMINAT SOLUTION OF LINEAR SYSTEMS

Students should be able to evaluate 2 x 2 and 3 x 3 determinant.
Determinant of a matrix is used to determine whether a system of equations has a single solution, and if so, find
the solution.
Evaluating (2 x 2) determinant:
a11 a12
 a11a22  a21a12
a21 a22
Practice:
y
8
3 1

6 2
2
 y 2  16  ( y  4)( y  4)
y
3 1

2 0
y 3

2 x
Evaluating (3 x 3) matrix
a11 a12
a21 a22
a31 a32
a13
a23  ( a11a22a33  a12a23a31  a13a21a32 )  ( a31a22a13  a32a23a11   a33a21a12 )
a33
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Practice on evaluating determinants:
3
2
1
0 
3
6
5 6
2
0 0 0
3 1 5 
6 7 9
4 2 1
3 0 5 
1 4 2
CRAMER’S RULE
For a system:
a1 x  b1 y  c1
a 2 x  b2 y  c2
The three determinants are:
D
a1
a2
b1
b2
Dx 
c1
c2
b1
b2
Dy 
a1
a2
c1
c2
If D = 0, Cramer’s rule does not apply – use other method.
The solution for the system is:
Dy
D
x x
y
D
D
Practice:
5 x  4 y  10
3 x  7 y  6
D
5 4
 94
3 7
Dx 
10 4
0
6 7
Dz 
5 10
3 6
x=2
y=0
Practice:
x yz4
2 x  y  3z  4
nx  ( n  1) y  ( n  2
( n  3) x  ( n  4) y  ( n  5)
4 x  2 y  z  15
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6.4. Partial Fractions
6.5. Nonlinear Systems of Equations
In this section student should be able to solve the system with on linear and one nonlinear equation (Fe. straight
line and parabola)
y  x2  2
Practice:
Use substitution method to find the interception and check your
x y0
work with graphing calculator.
Solution: Solve second equation for y and substitute in the first equation.
x  x2  2
x2  x  2  0
( x  2)( x  1)  0
x   2  y  2
x 1 y 1
( 2,2)
(1,1)
Graphical representation of the solution:
-4
-3
4
3
y=-x 2+2 2
1
0
-2
-1 -1 0
-2
-3
-4
x-y=0
1
2
3
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The coordinates of two points of intercept are the solution of the nonlinear system
Practice:
x  y 1
y  x2  1
y  ( x  3) 2
x  2 y  2
y  x2  4x
2 x  y  8
y  x 2  6x
3x  2 y  10
6.6. SYSTEM OF INEQUALITIES – LINEAR PROGRAMING
Solve the system of inequalities:
x  2 y  10
y  x2  1
The intersection of a solution sets of the two inequalities is the solution of the both inequalities. To obtain the
solution graph both inequalities in the same coordinate system:
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If P =2x + 3y, find the maximum value of P subject to given constrains:
x>0
y>0
x+y<4
2x + y < 6
Procedure:
1.
2.
3.
4.
5.
Graph each inequality
Find the region of feasible solution – region that contain solution for each inequality
Find the coordinates of vertices of the region of feasible solution
Calculate P for the coordinates of each vertices
One of the vertices will generate maximum P
6.7. Properties of Matrices

Student should be able to add, subtract, and multiply matrices manually and using graphing calculator.
Equivalent matrices – Corresponding elements are equal
x z
1 c

y 1
0 a


x  1
y0
zc
a 1
Adding two matrices – if they are of same size.
9 4
3 2
6
6


8 2
4 7
 12 9
If

3
A
4
and
1
B 2
3
Then A + B – never exist
Definition of Zero matrix – All elements are zero.
A+(-A)=0
3 1
 3 1
0 0


4 2
 4 1
0 0

Subtracting matrices – if they are of same size.
Practice: Solve for all variables:
2k  8 y
5k  6 y
 3k  2 y


6 z  3x
2 z  5x
4z  2x

2k  8 y  (5k  6 y )  3k  2 y
 Solve the system
6 z  3x  ( 2 z  5 x )  4 z  2 x
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Practice: Solve for all variables
x  2 1
2 x 10
10 9


 Generate two equations
2
y
2 8
4 5y
Multiplication of matrices

By the definition the product of (m x n) matrix and (n x p) matrix is said to exist because “the # of columns
in first matrix is equal to the # of rows in second matrix”
Matrix A
Matrix B
(3x2)
(2x 4)
Must match
3 x 4 is the size of A x B
Practice Q. 35 on page 501.
2 2 1
A
3 0 1
0 2
B  1 4
0 2
c11 c12
 2 10

c21 c22
0 8
A B 
c11  2  0  2  ( 1)  ( 1)  0  2
c21  3  0  0  ( 1)  1  0  0
c12  2  2  2  4  ( 1)  2  10
c22  3  2  0  4  1  2  8
6.8. MATRIX INVERSES

Identity matrix is square matrix with 1’s on a left-to-right diagonal and zero all other elements:
1 0 0
1 0
Example of 2 x 2 identity matrix:
and 3 x 3 identity
0 1 0
0 1
0 0 1

Multiplication property of identity of matrices:
matrix.
A I  A
INVERSE MATRIX A-1 is a matrix such: A  A1  I
and
where A is any matrix and I is identity
A1  A  I
A – matrix
A-1 – inverse of A
I – Identity matrix
Inverse matrix can be found for only square matrices (number of rows equal to number of columns).
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Procedure for finding inverse matrix:
TO obtain A-1 for any (n x n) matrix, for which A-1 exist, follow the steps:
1. Form an augmented matrix A I n  where In is (n x n) identity matrix.
2. Perform raw transformation on A I n  to get matrix of a form: I n B
3. Matrix B is A-1.
Example: Find the inverse of A:
A
6 4
3 2
3
2 3 1 0
1
1
R1( ) 
2
3 4 0 1
2
3 4
1 3
0
A1 
2
1
1
1
1 3
0
0
2
2
R
2

3

R
1

R 2 /(  1 ) 
2
2
3
1
0

1
0 1
2
2
1 0 4 3
0
2
R1  3 R 2 
2
0 1 3 2
3 2
1
4 3
3 2
Example of solving the system of equations using matrix method.
2 x  3 y  10
2 x  4 y  12
Matrix equation for the system:
2 3
x
 10


3 4
y
 12
A  X   B  X   A1  B
A1  ?
A1 
4 3
3 2
x  A1  B 
4 3
 10
4


3 1
 12
6
Solution of the system:
x = 4, and y = 6.
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