There are actually two reasons I am posting this question(Which has 3 parts)
1) Because it's really fun
2) Because I can't figure out parts 2 and 3.

Apply elimination to produce the factors L and U for


Part 1: A = \left[\begin{array}{cc}2&1\\8&7\end{array}\right]


Part 2: A = \left[\begin{array}{ccc}3&1&1\\1&3&1\\1&1&3\end{array}\r  ight]


Part 3: A = \left[\begin{array}{ccc}1&1&1\\1&4&4\\1&4&8\end{array}\r  ight]


Now for the first one, what I did was(and I hope I get the notation on this right)

E = Elimination Matrix,


EA = \left[\begin{array}{cc}1&0\\-4&1\end{array}\right] This produced L^-^1

Multiply L^-^1 by A, to get


U = \left[\begin{array}{cc}2&1\\0&3\end{array}\right]


A=LU since this brought me back to A, I knew I was right...
Now when getting into the 3x3 matrices, I started getting into trouble.
Do I basically use a guess and check method? Or is there a way I can systematically solve this using elimination matrices and LU?
Are Row Exchanges necessary for part 2 and 3?
Can I avoid using Guess and Check for L and U?
Also I'm going to put what I did in Part 1 in Matrix Notation(or at least attempt to) and was wondering if someone could tell me if it is correct.


E^-^1E_1A = U


Which would produce


E^-^1E_1A=E_1E^-^1U


L=E^-^1


U=E_1