Post on 30-Dec-2015
Umm… did she say 3 by 3??
Just like the 2 by 2 systems, we will solve the 3 by 3 systems.
How many methods did we choose from to solve the 2 by 2’s?
1. ________________________________
2. ________________________________
Conveniently, there are 3 methods we can use here
Today will be the first two
1. ________________________________2. ________________________________
Follow all directions, but unless specified you can use any method.
Guess which method would be best here:
1.
Yes – substitution is best for this type of problem. Specifically it is called “back substitution” because you substitute backwards from simplest to most complicated. However, substitution can be messy if there are multiple variables in each equation. Instead, you should probably
3x y z 3
7y 6z 13
5z 5
2 2
3 1
2 4 3 5
x y z
x y z
x y z
Steps to solve 3 Equations 3 Variables1. ___________________________________
2. ___________________________________
3. ___________________________________
4. ___________________________________
5. ___________________________________
OK – try it now and check.
2.
3.
x y z 1
x 3y z 7
4x y 3z 2
x y 2z 2
2x 3y z 3
3x 3y 4z 2(1, -1, -2)
(1, 2, 0)
One last thing:
What will you see with a no solution or what we used to call “infinite number of solutions?”
If you get something false (0 = 4) then there is no solution.
Method 3
This is called ____________________
It does involve more work in your head. You will be required to do a problem on the test using this method.
What is Matrix Elimination
Essentially you will take the coefficients from a 3 by 3, put it in a matrix and then work to make a new matrix by multiplying, dividing and/or combining rows.
Matrix – ________________________________
______________________________________
Lets work through a problem
1.
Now what we will do is create a new matrix.
2x y 3z 0
x y 2z 1
x 2y z 3
2 1 3 0
1 1 2 1
1 2 1 3
The ideal new matrix will have ones on the diagonal and 0’s to the left of the diagonal.
You may do this by a) ____________________________________b) ____________________________________c) ____________________________________
Lets try it together.
1 # # #
0 1 # #
0 0 1 #
So what did we do?We converted the original system
Into this:
which is
2x y 3z 0
x y 2z 1
x 2y z 3
2 1 3 0
1 1 2 1
1 2 1 3
1 1 2 1
0 1 1 2
0 0 1 1
Before we try it again.
Please realize this: just as different people will get different intermediate equations with regular elimination, you may not have the same matrix as someone next to you but it doesn’t mean you are wrong.
Also, just aim for the zeroes. In some problems getting the 1 or -1 on the diagonals is a mess, so I don’t look for that.
This is a great method.
Cramer’s Rule is a neat way to evaluate systems and if you put the work in now you’ll do fine. It can be used for any size (2 by 2, 3 by 3 or even larger) system.
It is easy to memorize and fast.
Definitions
Determinant – _______________________
2nd Order Determinant – _______________
3rd Order Determinant – _______________
Elements – _________________________
What does a determinant look like?
A 2nd order determinant looks like this
And the value of the determinant =______
____________ – ________________
a b
d e
1 1
2 2
3 3
1
2
3
b
a c
a
a b
c
b
c
1 1
2 2
3 3
1
2
3
b
a c
a
a b
c
b
c
1 1
2 2
3 3
1
2
3
b
a c
a
a b
c
b
c
x
1 1 1
2 2 2
3 3
1
3
2
3
a x b y c z
a x b y c z
a x b y c z
d
d
d
y z
Hint: if you can find x and y, just sub in to find z