# How do you find the #LU# factorization for #tildeA = [(1,0),(-3,1)]# such that #tildeL# is unit diagonal?

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I had gotten an #LU# factorization, except my #tildeL# matrix wasn't lower-triangular, so it doesn't count. Here's the expression I used:

If #tildeA = tildeLtildeU# , then when #tildeE_ncdotcdotcdottildeE_2tildeE_1tildeA -> tildeU# :

#color(blue)(tildeA = stackrel(tildeL)(overbrace(tildeE_1^(-1)tildeE_2^(-1)cdotcdotcdottildeE_n^(-1)))tildeU)# .

where #tildeE# is an elementary matrix and #tildeE^(-1)# is its inverse. #tildeU# is an upper-triangular matrix, and #tildeL# is a lower-triangular matrix.

I had gotten an

If

#color(blue)(tildeA = stackrel(tildeL)(overbrace(tildeE_1^(-1)tildeE_2^(-1)cdotcdotcdottildeE_n^(-1)))tildeU)# .where

#tildeE# is an elementary matrix and#tildeE^(-1)# is its inverse.#tildeU# is an upper-triangular matrix, and#tildeL# is a lower-triangular matrix.

##### 1 Answer

OH, I figured out what was going on. It turns out that the upper triangular matrix

So, I already have

#tildeL = [(1,0),(-3,1)]#

#tildeU = [(1,0),(0,1)]#

#tildeA = tildeLtildeU = [(1,0),(-3,1)][(1,0),(0,1)] = color(blue)([(1,0),(-3,1)])#

The longer way involves elementary matrices. Since each elementary row operation corresponds to the multiplication by an elementary matrix, we can do the following:

#stackrel(tildeA)(overbrace([(1,0),(-3,1)]))stackrel(3R_1 + R_2" ")(->)[(1,0),(0,1)] = tildeU#

Thus,

So what we have is:

#tildeE_1tildeA = tildeU#

the condition for which

As a result, we should realize that if we undo

#tildeE_1^(-1) = [(1,-3),(0,1)]#

#tildeItildeA = tildeA = tildeLtildeU = stackrel(tildeE_1^(-1))(overbrace([(1,0),(-3,1)]))stackrel(tildeU)(overbrace([(1,0),(0,1)])) = color(blue)(stackrel(tildeA)(overbrace([(1,0),(-3,1)])))#

so