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Hey, It's --OutpostMir--, bringing High School Geometry and Algebra student everywhere this nifty way to blaze right through linear equation problems using those wonderful bracketed sets we all call matrices!

Very Important Note: Be cool, and learn how to do these the hard way first before using this method. That way your teacher won't get pissed because you didn't do it their way.

Also, this assumes you have a TI-8x graphing calculator. If not, I'm sorry, but I can't help.

Here's the problem:
5x+3y=-1
-4x+2y-3

And here's how to solve:
1) First type 2nd and x-1. This takes you to the Matrix screen. Scroll over to "Edit" and hit Enter

2) First we need to define the size of the matrix. hit 2 and Enter and 2 and Enter. Enter the coefficients of the variables in the linear equations into the matrix. Once you are done hit 2nd and x-1 again to go back to the matrices screen.

3) Scroll over to Edit and chose the second menu choice. This time create a 2x1 matrix and fill it with the numbers on the other side of your equal sign. When your done, hit 2nd+mode to quit to the home screen.

4) The easiest part. Hit 2nd+x-1 and choose the first option on this screen. Hit the x-1 key,and then go back to the matrix menu and chose matrix B. Hit enter.

If you did everything right, the values for X (top number) and Y (bottom number) will automagicly appear on screen! Huzzah! You did it!

Other (sometimes easier) ways to solve multivariate equations with matrices:
The equation used will be as follows:

 x + 3y + 2z =  7
2x -  y      = -3
-x + 2y + 5z = 12

Its coefficient matrix would be:

[ 1  3 2]
[ 2 -1 0]
[-1  2 5]


On TI-8x calculators, you can edit matrices in the matrix editor, then paste the variable to the screen, or you can enter them inline, like so:

[3 2]
[4 7]

would be entered in as:

[[3,2][4,7]]

Cramers Rule:

Set up the coefficient matrix. Take the determinant of that matrix. Each variable will be equal to a numerator divided by that determinant. To find the numerator, replace the column for that variable with the constants in the original equations.

Example:

    | 7  3 2|
    |-3 -1 0|
    |12  2 5|    22
x = --------- = --- = -22/29
    | 1  3 2|   -29
    | 2 -1 0|
    |-1  2 5|

    | 1  7 2|
    | 2 -3 0|
    |-1 12 5|   -43
y = --------- = --- = 43/29
    | 1  3 2|   -29
    | 2 -1 0|
    |-1  2 5|

    | 1  3  7|
    | 2 -1 -3|
    |-1  2 12|   -48
z = ---------- = --- = 48/29
    | 1  3  2|   -29
    | 2 -1  0|
    |-1  2  5|


Reduced Row-Echelon Form:

This is even easier: you just use the rref() function on your calculator, and input the coefficient matrix with the constants appended as another column.

[ 1  3 2  7]
[ 2 -1 0 -3]
[-1  2 5 12]

is your matrix. You just use rref(matrix) so if you wanted to do this inline, you would enter: rref([[1,3,2,7][2,-1,0,-3][-1,2,5,12]]) and it would spit out something like:

[1 0 0 -.7586206897]
[0 1 0 1.482758621 ]
[0 0 1 1.655172414 ]

which you could use the handy >FRAC command to get:

[1 0 0 -22/29]
[0 1 0 43/29 ]
[0 0 1 48/29 ]
Or, if you don't have a flashy calculator...

Given a system of linear equations:

a1,1x1 + a1,2x2 ... + a1,n = b1
a2,1 + ... = b2
.
.
.
am,1 + ... = bm

We can represent this system as a matrix equation:

A . X = B

Where A is the m*n coefficient matrix, whose (i,j) entry is ai,j,
X is [ x1, x2, ... , xn ] ,
and B is [ b1, b2, ... , bm ]


Then, we define the augmented matrix of this system to be [ A|B ]
(i.e. an m*(n+1) matrix given by adding B as an extra column to the right side of A)

This matrix can now be transformed, using elementary row operations, into row reduced echelon form.

Given a row-reduced matrix E, we can then use the Gauss-Jordan procedure to find solutions.

Case 1: The last non-zero row of E is 0,0,...,1. In this case, the system is inconsistent.

Case 2: E has n non-zero rows. Then (x1 , x2, ... , xn) = the rightmost column of E.

Case 3: E has k non-zero rows. Assign parametric values to the n-k variables whose rows in E are zero.

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