Cramer’s Rule

Cramer’s Rule is a method of solving systems of equations. There are many methods including substitution and elimination. Sometimes though the coefficients of the equation are ugly. (Ugly isn’t my terminology, it is one given to me by a student.) To many students anything that isn’t a small whole number is ugly. So, when a system of equations includes fractions or decimal numbers, Cramer’s Rule has a way of simplifying what might be an otherwise cantankerous solution method.

Cramer’s Rule involves setting up something called a determinant using the coefficients (numbers before the variables).

Given the following system of equations:

Where x and y are variables and A, B, C, D, E, F are constants

Ax + By = E
Cx + Dy = F

The determinants are placed in a fraction to solve for each variable X and Y

It is written as

FOR X:

|E  B |
| F D |
_____
| A B |
| C D |

 

FOR Y:

| A E |

| C F |
_____
| A B |
| C D |

As you might have noticed, the denominators are the same and in alphabetical order for X and Y. That makes the denominator pretty easy to remember, but what about the numerators????

Well, mnemonic devices are memory techniques that utilizes acronyms or sentences or even music to help a student remember important items. One of my latest favorites is a method to remembering Cramer’s Rule.

My mnemonic for Cramer’s Rule is EVERY BOY FINDS DOGS for X and ALL ELEPHANTS CAN’T FORGET for Y. I had a student change the Y to ALL ELEPHANTS CAN’T FLY. Whatever works for you is the one to use.

Anyhow, so what does the alphabet soup tell you once you have remembered it???

Well, after you have set up the determinants, it is just arithmetic from here on in.

For X = (E*D – B*F) divided by (A*D – B*C)
And Y = (A*F – E*C) divided by (A*D – B*C)

 

 

Here is one example:

2x-3y = 9
x+5y =-2

FOR X

|9   -3|
|-2  5|                                      (9 * 5) – (-3 * -2)                           45 – 6               39
____         =                         ____________        =                        ______      =    __       =       3
|2  -3|                                      (2 * 5) – (-3 * 1)                            10 – -3               13
|1    5|

 

FOR Y

|2   9|
|1  -2|                                     (2 * -2) – (9 * 1)                                   -4 – 9                 -13
____          =                       ___________             =                           ____        =        __      =       -1
|2  -3|                                     (2 * 5) – (-3 * 1)                                  10 – -3                13
|1    5|

 

 

So, the solution set for the above example is (3, -1)

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