You may or may not have seen this before -- it depends on where you took your last Algebra class.

First, I need to tell you about determinants. We'll be using these to solve systems.

Say you've got a 2 x 2:

x - 3y = 4

5x + 7y = 8

We can make something called a "coefficient matrix":

a coefficient matrix ... [ top row: 1 , -3 bottom row: 5 , 7 ]

A matrix is just a grid of numbers with brackets around them. (Remember that a coefficient is the number in front of the variable... x = 1x, so the coefficient is 1.) (We'll learn more about these matrix things in the next chapter.)

We've got rows and columns of the matrix:

[ top row: 1 , -3 bottom row: 5 , 7 ] ... row 1 = 1 , -3 ... row 2 = 5 , 7 ... column 1 = 1 , 5 ... column 2 = -3 , 7

A determinant (which uses vertical lines instead of brackets)

| row 1: 1 , -3 row 2: 5 , 7 |

will give us a number that goes along with the matrix. We'll be able to use these numbers to solve systems! Woo hoo!

OK, so here's how you get the determinant -- it's really easy:

| row 1: 1 , -3 row 2: 5 , 7 | = ( 1 ) ( 7 ) - ( 5 ) ( -3 ) = 22
Here's what I did:

I multiplied down this diagonal...

| top row: 1 , -3 bottom row: 5 , 7 | ... multiply down diagonally through the 1 and the 7 = ( 1 ) ( 7 )
| top row: 1 , -3 bottom row: 5 , 7 | = ( 1 ) ( 7 ) - Put a minus...
then multiplied up this diagonal...

| top row: 1 , -3 bottom row: 5 , 7 | ... multiply up diagonally through 5 and -3 = ( 1 ) ( 7 ) - ( 5 ) ( -3 ) = 22


Here's the general formula:

| top row: a , b bottom row: c , d | = ad - cb


TRY IT:

| top row: -2 , -6 bottom row: 5 , 8 | =

| top row: 3 , 9 bottom row: 2 , 6 |

Cramer's Rule uses determinants to solve systems and was named after the wacky guy on Seinfeld. (OK, I made that last part up.)

Let's just do one and I'll show you how it works:

3x - y = 7

-5x + 4y = -2

First, we'll get the determinant of the coefficient matrix -- we'll call it D:

D = | top row: 3 , -1 bottom row: -5 , 4 | = ( 3 ) ( 4 ) - ( -5 ) ( -1 ) = 7
Now, we're going to find two more determinants.

The first one we'll call Dx -- here's how it goes:
Take D...

| top row: 3 , -1 bottom row: -5 , 4 |
and delete the column for the x guys...

| top row: _ , -1 bottom row: _ , 4 | ... column 1 are the x guys and column 2 are the y guys
Replace that column with the " = guys " (the 7 and the -2) and you get

Dx = | top row: 7 , -1 bottom row: -2 , 4 | = ( 7 ) ( 4 ) - ( -2 ) ( -1 ) = 26


To get the x part of our (x, y) solution, we take

x = ( Dx / D ) = ( 26 / 7 )

Now, to get the y part...

Take D again...

| top row: 3 , -1 bottom row: -5 , 4 |
and delete the column for the y guys...

| top row: 3 , _ bottom row: -5 , _ | ... column 1 are the x guys and column 2 are the y guys
Replace that column with the "= guys":

Dy = | top row: 3 , 7 bottom row: -5 , -2 | = ( 3 ) ( -2 ) - ( -5 ) ( 7 ) = 29
So, our y part is

y = ( Dy / D ) = ( 29 / 7 )



and our final answer is ( 26 / 7 , 29 / 7 ) . That's it!


YOUR TURN:

Use Cramer's Rule to solve

-6x + 8y = 17

13x - 2y = -4

I really think that Cramer's Rule is the easiest method for solving 2 x 2's. Especially if the answer is messy fractions (and it usually is)!

There's one little glitch...

Check this guy out:

7x - 7y = 8

-3x + 3y = 2

D = | top row: 7 , -7 bottom row: -3 , 3 | = ( 7 ) ( 3 ) - ( -3 ) ( -7 ) = 0

Whatever Dx and Dy are, we'll be stuck with this:
x = ( Dx / 0 ) and y = ( Dy / 0 ) ... Zeros in denominators are bad news!

What does this mean? Well, we aren't going to get a nice (x, y) answer... In fact, we won't get ANY (x, y) answer! Either the lines are parallel or they are the same line. The only problem is that we can't tell. This is when you'll want to fall back on elimination to see which it is.

Which is it?