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.
A = LU
I11 0 0
I21 I22 0
I31 I32 I33
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...