When you are given a matrix and expected to find the X, Y, and Z value, you can use the computer and have it do it for you, lazy, or you can do it by hand and show the world that you can do it!

Here is the way I do it. Gaussian Elimination.

Suppose we have this matrix, lets call it matrix M.

1 -3 1 4

2 -8 8 -2

-6 3 -15 9

Never fear, We'll just use the good ol' Gaussian Elimination and show this matrix who's boss.

First I started off by saying I wanted to get this into this form

1 0 0 0

0 1 0 0

0 0 1 0

The first thing I did was got the first row and used it to zero out M(1,3)

R3=R3- (-6*R1)

1 -3 1 4

2 -8 8 -2

0 -15 -9 33

We then need to get M(1,2) to zero out.

R2=R2 -(2*R1)

1 -3 1 4

0 -2 6 -10

0 -15 -9 33

Now we can divide rows 2 and 3 by -2 and -3 respectively.

R2=R2/-2

R3=R3/-3

1 -3 1 4

0 1 -3 5

0 5 3 -11

Lets further refine R3

R3=R3 - (5*R2)

1 -3 1 4

0 1 -3 5

0 0 18 -36

R3=R3/18

1 -3 1 4

0 1 -3 5

0 0 1 -2

Now as you can see there is a solvable equation but I don't feel like solving this, lets keep going unless your lazy then just solve it I suppose. :p

Lets see what happens when you further refine R1

R1=R1+(3*R2)

1 0 -8 19

0 1 -3 5

0 0 1 -2

Not done yet, where you going?

R1=R1+(8*R3)

1 0 0 3

0 1 -3 5

0 0 1 -2

And finally…….R2

R2=R2+(3*R3)

1 0 0 3

0 1 0 -1

0 0 1 -2

Ok now we can go ahead and solve…….wait……oh nevermind. Its already solved. Wonder how the boys and girls that stopped further up the page are doing?

Either way, we solved it and now I want Wingstop.

*Runs off to Wingstop*