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## Homework Statement

Let

A =

4 -4 -1

4 -4 0

4 0 0

Use the criterion from the lectures to show that A does not have an LU factorisation such that A = LU with lower and upper triangular matrices L and U.

How can you re-arrange the matrix rows such that it is evident without much computation that the permuted matrix does have an LU factorisation.

## Homework Equations

A = LU

L =

I11 0 0

I21 I22 0

I31 I32 I33

U =

U11 U12 U13

0 U22 U22

0 0 U33

## The Attempt at a Solution

I solved part 1, but taking a 2x2 sub-matrix and showing that its determinant is zero. [-4 0 ; 0 0] ----> det =0

The second part where I have to show that A has an LU factorisation, I merely took A = LU and expanded upon that...

4 -4 -1 = I11 0 0 U11 U12 U13

4 -4 0 I21 I22 0 x 0 U22 U23

4 0 0 I31 I32 I32 0 0 U33

= I11U11 I11U12 U13

I21U11 I21U12 + I22U22 I21U13 + U22

I21U11 I31U12 + I32U22 I31U13 + I32U23 + U33

Let I11 = I22 = I33 = 1

= U11 I11U12 U13

I21U11 I21U13 + U22 I21U13 + U22

I31U11 I31U12 + I32U22 I31U13 + I32U23 + I33U33

Substituting the values of A into the LU, we get

=

U11 = 4 I11U12 = -4 U13 = -1

I21U11 = 4 I21U13 + U22 = -4 I21U13 + U22 = 0

I31U11 = 4 I31U12 + I32U22 = 0 I31U13 + I32U23 + I33U33 = 0

Now at this form, it's becoming obvious that row 1 and row 3 needs to be swapped. There is a single pivot on the left-most side of the third row, and that'll allow for an LU factorisation. Unfortunately, I can't really rationalise my position outside of that...