Question

would someone explain me how do I get the L matrix in the LU factorization?
i've got an example if necessary:
2x-y+3z=1
-4x+5y=-2
4x+2y+18z=0

Answer 1

it will take too many writings, so i'll explain one step
by adding 2 times of row 1 to row 2, you get 3y+6z = 0

now 2 times of row 1 to row 2 is actually multiplying the following matrix
1 0 0
2 1 0
0 0 1

so you get
row subtract matrix * original matrix = result matrix
1 0 0 2 -1 3 1 2 -1 3 1
2 1 0 * -4 5 0 -2 = 0 3 6 0
0 0 1 4 2 18 0 4 2 18 0

you keep multiplying matrix until right hand side matrix is U (upper triangle matrix)
At the end,
Lower trianglar matrix * original matrix = Upper trianglar matrix.
Now you multiply Lower trianglar matrix's inverse to both side
original matrix = Lower's inverse * U
Lower's inverse is actually also another Lower trianglar matrix.(we call this L)
So you get A = L*U

Answer 2

thank you ^^

Answer 3

Pcompassion's answer is perfectly correct, but if I understand Professor Strang correctly, there's a second way to get to L.

Let's say you did the following steps (this is not your example):
1) add 2x row_1 to row_2
2) substract 3xrow_2 to row_3

Then L is:
\[\left[\begin{matrix}1 & 0 & 0\\ 2 & 1 & 0 \\ 0 & -3 & 1\end{matrix}\right]\]

Answer 4

Pcompassion's method is ok but it ignores the basic idea of using the LU decomposition, it is used in the practical calculations beacause LU deomposition can be calculated directly! with out solving the equations simultaniously , for example is the Matrix A is


A = ⎡ ⎤
| a11 a12 a13 |
| a21 a22 a23 |
| a31 a32 a33 |
⎣ ⎦
then
⎡ ⎤ ⎡ ⎤
| 1 0 0 | | u11 u12 u13 |
L xU= | m21 1 0 | X | 0 u22 u23 |
| m31 m32 1 | | 0 0 u33 |
⎣ ⎦ ⎣ ⎦
Based on the requirement of the diogonal 1's in L matrix this mathod is called Doolittle's Method .
now if you multiply first row of L to first column of U and equate it to the a11 of you will get the u11, then to the 2nd solumn and equate it to a12 u will get u12 as u11 is already known and so on !

depending on this method following algorithem can be used for any size of the matrix to decompose it in to LU factorization requiring the less no. of operations

\[u _{1k} = a _{1k} \]\[ k=1 . . n\]
\[m _{j1} = a _{j1}/u _{11 } \] \[ j= 2 . . n\]
\[u _{jk}= a _{jk}-\sum_{s=1}^{j-1}m _{js}u _{sk}\]
\[k = j.....n;j \ge 2\]
\[m _{jk}= 1/u _{kk }\left( a _{jk}-\sum_{s=1}^{k-1}m _{js}u _{sk} \right) \]
\[j=k+1...n; k \ge 2\]
so it become really easy to calculate LU factorization