Download Unit10 PowerPoint for trigonometry class

Welcome
to Week 10
MA1310
College Math 2
Systems of Equations
A set of two equations
containing two variables ...
A set of three equations
containing three variables ...
Are you seeing a pattern?
Systems of Equations
2x + 5y = 38
x + 7y = 37
Lots of equations come in groups!
Systems of Equations
If they give you values for “x”
and “y” and ask if they are
true values, plug them in to
both equations – both equations
must be true for the values
Systems of Equations
IN-CLASS PROBLEM
Are x = 4 and y = 6 true
values for the system:
2x + 5y = 38
x + 7y = 37
Plug in x = 4 and y = 6
Systems of Equations
IN-CLASS PROBLEM
Plug in x = 4 and y = 6
2(4) + 5(6) =? 38
(4) + 7(6) =? 37
Are the values x = 4 and y = 6
a true solution to this system?
Systems of Equations
To solve a system, you must
have at least as many equations
as you have variables
Systems of Equations
IN-CLASS PROBLEM
Suppose we wanted to find “x”
and “y” for the equations:
2x + 5y = 38
x + 7y = 37
Systems of Equations
IN-CLASS PROBLEM
Strategy: try to get one of the
variables written in terms of
the other variable:
2x + 5y = 38
x + 7y = 37
Look for a “loner”
Systems of Equations
IN-CLASS PROBLEM
x + 7y = 37
Use algebra to solve for “x”:
x = 37 – 7y
Now, plug in this “x” into the
other equation
Systems of Equations
IN-CLASS PROBLEM
2x + 5y = 38
x = 37 – 7y
To get:
2(37-7y) + 5y = 38
Now, solve for “y”!
Systems of Equations
IN-CLASS PROBLEM
2(37-7y) + 5y = 38
74 – 14y + 5y = 38
-9y = 38 – 74 = -36
y = -36/-9
y = 4
Then we plug in this value for
“y” and solve for “x”
Systems of Equations
IN-CLASS PROBLEM
We already have:
x = 37 – 7y
y = 4
So just plug in y = 4:
x = 37 – 7(4)
x = 37 – 28
x = 9
Systems of Equations
IN-CLASS PROBLEM
So, the true solution to the
system:
2x + 5y = 38
x + 7y = 37
is x = 9, y = 4
or (x,y) = (9,4)
Systems of Equations
IN-CLASS PROBLEM
To find the true solution to the
system:
2x + 5y = 38
3x + 7y = 50
Solve one of the equations for
either “x” or “y”
Systems of Equations
IN-CLASS PROBLEM
2x + 5y = 38
2x = 38 – 5y
x = 38 – 5y
2
Then plug in this “x” into the
other equation
Systems of Equations
IN-CLASS PROBLEM
3x + 7y = 50
x = 38 – 5y
2
Systems of Equations
IN-CLASS PROBLEM
3(38/2 - 5y/2) + 7y = 50
57 – 7.5y + 7y = 50
Solve for “y”:
-7.5y + 7y = 50 – 57
-0.5y = -7
y = -7/-0.5
y = 14
Systems of Equations
IN-CLASS PROBLEM
You already have:
x = 38 – 5y
2
and
y = 14
so:
x = 38 – 5(14) = 38-70
2
2
x = -32/2 = -16
Systems of Equations
IN-CLASS PROBLEM
So the true solution to the
system:
2x + 5y = 38
3x + 7y = 50
is x = -16, y = 14
or (x,y) = (-16,14)
Systems of Equations
For more equations –
Same stuff, just more of it!
Systems of Equations
IN-CLASS PROBLEM
Solve the following system of
equations:
x + z = 3
x + 2y – z = 1
2x – y + z = 3
Systems of Equations
IN-CLASS PROBLEM
x + z = 3
is the easiest – let’s start there:
z = 3 – x
Plug that in!
Systems of Equations
IN-CLASS PROBLEM
z = 3 – x
x + 2y – z = 1
2x – y + z = 3
become:
x + 2y – (3 – x) = 1
2x – y + (3 – x) = 3
Now it’s just two equations!
Systems of Equations
IN-CLASS PROBLEM
x + 2y – 3 + x = 1
2x – y + 3 – x = 3
or
2x + 2y = 4
x – y = 0
Use x – y = 0  x = y
Systems of Equations
IN-CLASS PROBLEM
x = y
2x + 2y = 4
2x + 2x = 4
4x = 4
x = 1
Systems of Equations
IN-CLASS PROBLEM
So,
x =
x =
z =
now we have:
1
y
3 – x
So, (x,y,z) = (1,1,2)
Uses for Systems of Equations
Lots of business applications
use systems of equations
Uses for Systems of Equations
The income formula is the total
amount generated by selling the
product:
I(x) = (price per unit sold) * x
x is the quantity sold
Uses for Systems of Equations
The cost formula is the total
cost of producing the product:
C(x) = fixed cost +
(cost per unit produced) * x
x is the quantity produced
Uses for Systems of Equations
The profit formula is the
difference between the cost and
the revenue formulas:
P(x) = I(x) – C(x)
x is the quantity of the product
Uses for Systems of Equations
The x-solution for the two
equations Cost and Income
is the true solution of the
system of the two equations
(the Cost and Income equations)
Uses for Systems of Equations
The solution to the system is
called the “break-even point”
Uses for Systems of Equations
Video
Revenue = 25x – 0.01x2
Cost = 1600 + 5x + 10x
Systems of Equations
IN-CLASS PROBLEM
Revenue = 25x – 0.01x2
Cost = 1600 + 5x + 10x
Systems of Equations
IN-CLASS PROBLEM
Example:
A widget-maker produces
widgets
Each widget sells for $300
They sell 20,000 widgets each
year
What is their annual income?
Systems of Equations
IN-CLASS PROBLEM
I(x) = (price per unit sold) * x
I = $300 (20,000)
= $6,000,000
Systems of Equations
IN-CLASS PROBLEM
They have a fixed cost (rent,
utilities, salaries, etc.) of
$950,000 per year
Each widget costs $10 to make
They make 20,000 widgets
each year
What is their annual cost?
Systems of Equations
IN-CLASS PROBLEM
C(x) = fixed cost +
(cost per unit produced) * x
C = $950,000 + $10(20,000)
= $1,150,000
Systems of Equations
IN-CLASS PROBLEM
What is their annual profit?
Systems of Equations
IN-CLASS PROBLEM
P(x) = I(x) – C(x)
P = $6,000,000 - $1,150,000
= $4,850,000
Systems of Equations
IN-CLASS PROBLEM
Let’s buy Widget stock!!!
Systems of Equations
IN-CLASS PROBLEM
Break-even point:
How many widgets you need to
make for the cost to equal the
income (profit of $0)
Systems of Equations
IN-CLASS PROBLEM
Break-even point:
C(x) = I(x)
fixed cost + (cost per unit) * x
=
(price per unit sold) * x
Systems of Equations
IN-CLASS PROBLEM
Break-even point:
950,000 + 10(x) = 300(x)
Algebra magic…
950,000 = 300x - 10x
950,000 = 290x
950,000/290 = x
So x = 3275.86 units
Systems of Equations
IN-CLASS PROBLEM
They will “break even” when
they make and sell 3276 units
Since they make and sell a lot
more, they are making a huge
profit!
Uses for Systems of Equations
If you graph the income and
cost curves, the break-even
point is
where the
two curves
cross
Uses for Systems of Equations
Actually this is true for all
systems of equations – the true
solution (x,y) is
the point where
the graph of the
two curves cross
Uses for Systems of Equations
IN-CLASS PROBLEM
What is the break-even point?
Questions?
Matrices
Matrix (plural matrices) - a
rectangular table of elements
(or entries), numbers or
abstract quantities
Arranged in rows and columns
Matrices
Matrices in everyday life
Matrices
The term "matrix" for
arrangements of numbers was
introduced in 1850 by James
Joseph Sylvester
Matrices
Horizontal lines in a matrix are
called rows
Vertical lines are called columns
Matrices
A matrix with m rows and n
columns is called an m-by-n
matrix (written m × n)
m and n are called its dimensions
Matrices
The dimensions of a matrix are
always given with the number of
rows first, then the number of
columns:
r x c
Matrices
Matrices are usually given
capital letter names like “A”
Matrices
To designate a specific element
in a matrix, you use small
letters with subscripts r and c:
a37 means the element in matrix
A in row 3 column 7
Matrices
In Math class, matrices are
shown by listing the elements
inside square braces: [ ]
Matrices
┌
│ a11 a12 a13
A = │ a21 a22 a23
│ a31 a32 a33
└
┐
│
│
│
┘
Matrices
The Matrix Game
Questions?
Matrices
A matrix can be used as a
shorthand way of writing a
system of equations
Matrices
Called an augmented matrix:
Each row is an equation
Each column is the coefficient
of a variable or a constant
A vertical bar separates
coefficients on the left side
from constants on the right
Matrices
For example:
3x + 7y – 4z = 10
2x + 5y + z = 9
x + 6z = 15
Matrices
3x + 7y – 4z = 10
2x + 5y + z = 9
x +
6z = 15
┌
eq 1 │
eq 2 │
eq 3 │
└
x
3
2
1
y
z
becomes:
k
┐
7 -4 │ 10│
5 1 │ 9│
0 6 │ 15│
┘
Matrices
It is traditional to use “k” to
designate the constant rather
than “c”
(Don’t ask me why!)
Matrices
IN-CLASS PROBLEM
What would be the matrix for
the system of equations:
46x – 51y = 709
30x + 88y = -420
Matrices
IN-CLASS PROBLEM
46x – 51y = 709
30x + 88y = -420
x
y
k
┌
┐
eq 1 │ 46 -51 │ 709│
eq 2 │ 30 88 │-420│
└
┘
Matrices
IN-CLASS PROBLEM
What would be the matrix for
the system of equations:
x = 7y + 8
y = x – 3
Careful!
Matrices
IN-CLASS PROBLEM
x = 7y + 8
y = x – 3
x
x – 7y = 8
-x + y = -3
y
k
┌
┐
eq 1 │ 1 -7 │ 8 │
eq 2 │ -1 1 │-3 │
└
┘
Matrices
IN-CLASS PROBLEM
You can also recover the
original equations from a matrix
What would be the equations
for the matrix:
x
y
k
┌
┐
eq 1 │ 10
4 │ 92│
eq 2 │-43
0 │ 87│
└
┘
Matrices
x
y
k
┌
┐
eq 1 │ 10
4 │ 92│
eq 2 │-43
0 │ 87│
└
┘
The equations would be:
10x + 4y = 92
-43x = 87
Matrices
The goal of a system of
equations is to solve the system
for values of the variables that
work in all of the equations in
the system
Matrices
Same thing with a matrix!
Matrices
IN-CLASS PROBLEM
How would you solve:
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
1
0
0
0
1
0
z
k
┐
0 │ 10│
0 │ 9│
1 │ 15│
┘
Matrices
IN-CLASS PROBLEM
How would you solve:
┌
eq 1 │
eq 2 │
eq 3 │
└
x = 10,
x
1
0
0
y
y
z
k
┐
0 0 │ 10│
1 0 │ 9│
0 1 │ 15│
┘
= 9, z = 15
Matrices
This pattern is easy to solve:
upper triangle
┌
┐
eq 1 │ 1 0 0 │ 10│
eq 2 │ 0 1 0 │ 9│
eq 3 │ 0 0 1 │ 15│
└
┘
lower triangle
Matrices
IN-CLASS PROBLEM
What if you had:
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
5
0
z
k
┐
2 │ 19│
4 │ 15│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
This is called a “lower
triangular matrix”
x y z
k
┌
┐
eq 1 │ 6 3 2 │ 19│
eq 2 │ 0 5 4 │ 15│
eq 3 │ 0 0 1 │ 5│
└
┘
Matrices
IN-CLASS PROBLEM
Not quite as easy!
What value do you know?
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
5
0
z
k
┐
2 │ 19│
4 │ 15│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
5
0
z
k
┐
2 │ 19│
4 │ 15│
1 │ 5│ z = 5 !
┘
Matrices
IN-CLASS PROBLEM
We’ll use the same techniques
you used in solving systems of
equations to solve for y and x
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
5
0
z
k
┐
2 │ 19│
4 │ 15│
1 │ 5│
┘
Matrices
Remember - each row in an
augmented matrix represents an
equation
So…anything that you can do to
an equation can be done to the
row of a matrix!
Matrices
Scalar multiplication:
3x + 7y = 15
2(3x + 7y) = 2(15)
You can multiply a row of an
augmented matrix by a constant
Matrices
Scalar multiplication uses a
special notation:
2R1
This means multiply all of the
elements in row 1 by the
constant value 2
Matrices
Just like you can add two
polynomial equations, you can
also add two rows of an
augmented matrix:
3x + 7y = 15
5x - 2y = 40
8x + 5y = 55
Matrices
You can multiply by a constant
and add rows of a matrix at
the same time
Matrices
3R1 + R2
means multiply the elements in
row 1 by three, add it to the
elements in row 2, and replace
the elements in ROW 2 with
the result
Matrices
IN-CLASS PROBLEM
So… back to our problem:
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
5
0
z
k
┐
2 │ 19│
4 │ 15│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
If you multiply row 3 by -4 and
add it to row 2, you can solve
for y easily!
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
5
0
z
k
┐
2 │ 19│
4 │ 15│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
┌
eq 1 │
eq 2 │
eq3*-2
new eq2
eq 3 │
└
x
y
z
6
0
0
0
0
3 2
5 4
0 -4
5 0
0 1
k
┐
│ 19│
│ 15│
-20
-5
│ 5│
┘
Matrices
IN-CLASS PROBLEM
x
y
┌
eq 1 │ 6 3
eq 2 │ 0 5
eq 3 │ 0 0
└
Now what?
z
k
┐
2 │ 19│
0 │ -5│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
x
┌
eq 1 │ 6
eq 2 │ 0
eq 3 │ 0
└
Divide row
y
z
k
┐
3 2 │ 19│
5 0 │ -5│
0 1 │ 5│
┘
2 by 5!
Matrices
IN-CLASS PROBLEM
x
y
z
┌
eq 1 │ 6 3 2 │
eq 2 │ 0 1 0 │
eq 3 │ 0 0 1 │
└
Now what values do
k
┐
19│
-1│
5│
┘
you know?
Matrices
IN-CLASS PROBLEM
So now we know 2 values:
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
1
0
z
k
┐
2 │ 19│
0 │ -1│ y = -1
1 │ 5│ z = 5
┘
Matrices
IN-CLASS PROBLEM
Use the same strategy to solve
for x!
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
3
1
0
z
k
┐
2 │ 19│
0 │ -1│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
-3R2 + R1:
┌
eq 1 │
x
6
+0
6
eq 2 │ 0
eq 3 │ 0
└
y
z
k
┐
3 2 │ 19│
-3 +0
3
0 2
22
1 0 │ -1│
0 1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
Now what?
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
6
0
0
0
1
0
z
k
┐
2 │ 22│
0 │ -1│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
-2R3 + R1:
┌
eq 1 │
x
y
z
6 0 2
+0 +0 -2
6 0 0
eq 2 │ 0 1 0
eq 3 │ 0 0 1
└
k
┐
│ 22│
-10
12
│ -1│
│ 5│
┘
Matrices
IN-CLASS PROBLEM
x
y
┌
eq 1 │ 6 0
eq 2 │ 0 1
eq 3 │ 0 0
└
What now?
z
k
┐
0 │ 12│
0 │ -1│
1 │ 5│
┘
Matrices
IN-CLASS PROBLEM
x
┌
eq 1 │ 6
eq 2 │ 0
eq 3 │ 0
└
Divide row
y
z
k
┐
0 0 │ 12│
1 0 │ -1│
0 1 │ 5│
┘
1 by 6!
Matrices
IN-CLASS PROBLEM
x
y
z
┌
eq 1 │ 1 0 0 │
eq 2 │ 0 1 0 │
eq 3 │ 0 0 1 │
└
Now what values do
k
┐
2│
-1│
5│
┘
we know?
Matrices
IN-CLASS PROBLEM
Now what do we know?
┌
eq 1 │
eq 2 │
eq 3 │
└
x
y
1
0
0
0
1
0
z
k
┐
0 │ 2│ x = 2
0 │ -1│ y = -1
1 │ 5│ z = 5
┘
Matrices
So… if you have an upper and
lower triangular matrix, the
solution is obvious:
upper triangle
┌
┐
eq 1 │ 1 0 0 │ 10│
eq 2 │ 0 1 0 │ 9│
eq 3 │ 0 0 1 │ 15│
└
┘
lower triangle
Matrices
And… if you have only one zero
triangle, it’s more work:
┌
┐
eq 1 │ 1 1 2 │ 19│
eq 2 │ 0 1 0 │ 3│
eq 3 │ 0 0 1 │ 5│
└
┘
lower triangular matrix
Matrices
The upper and lower triangular
form is called the “GaussJordan” solution, the lower
triangular form is called the
“Gauss” solution
Matrices
Carl Friedrich Gauss
Wilhelm Jordan
Matrices
Wilhelm Jordan
Camille Jordan
Matrices
Computers and calculators
manipulate the matrix to
achieve the Gauss-Jordan
solution
Matrices
Computers and calculators do
these processes using a
strategy to reduce the matrix
to the Gauss-Jordan “Easy-ToSolve” form
Excel Solver Demo
Suppose you had the system of equations:
7u + 6v + 15w – 44x + 18y + 11z = 70
9u + 25v + 10w + 3x – 14y + 8z = 161
13u + 8v + 2w + 6x + 62y + 15z = 671
–2u +
v –
5w + 57x + 9y – 3z = 354
5u – 17v + 18w + 9x – 7y + 5z = 82
4u + 3v + 16w + 14x + 2y + 7z = 258
Solve for u, v, w, x, y, z
Questions?
Liberation!
Be sure to turn in your
assignments from last week
to me before you leave
Don’t forget
your homework
due next week!
Have a great
rest of the week!