9-1 Solving 3 by 3 Systems

Day 1

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.

What if you get a True Statement?

In this situation, the planes intersect at a line.

9-1 Solving 3 by 3 Systems

Day 2

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 #

2 1 3 0

1 1 2 1

1 2 1 3

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.

2. 2x 14y 4z 2

4x 3y z 8

3x 5y 6z 7

1 7 2 1

0 1 5 14

0 26 12 10

9-4 Determinants and Cramer’s Rule

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

Examples

Evaluate

1.

2. 2 2

6 1

3 1

1 3

Cramer’s Rule

Given a system

ax by c

dx ey f

6x 7y 9

x y 5

5x 4b 1

2x b 10

9-5 Higher Order Determinants

Cramer’s Rule and 3 by 3’s

How to Evaluate a 3 by 3 Determinant

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

3 1 2

1 0 3

4 2 1

Example

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

Examples

2.

x 2y z 3

2x y z 4

x y 2z 5 (2,1,-1)