Question

suppose we have $A = \begin{bmatrix} 0& 1 & 3 & 0\\ 0& -1 &-2 &3 \\ 1& -2 & 1 &0 \\ 2& -4 &4 &1 \end{bmatrix}$

a) find a factorization of A into the product MU where U is upper triangular (that is, find M and U such that A = MU where U is upper triangular).

b) find a permutation matrix P such that PA = LU where L is a lower triangular matrix and U is the same upper triangular matrix found in part a).

Answer 1

IA = A [1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1] A = [0 0 1 2 1 -1 -2 -4 3 -2 1 4 0 3 0 1] R_1 rightleftarrow R_3 p leftarrow [0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1] A = [1 0 0 2 -2 -1 1 -4 1 -2 3 4 0 3 0 1] so, here p = [0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 1] where p = permutation matrix now find LU factorization [1 0 0 2 -2 -1 1 -4 1 -2 3 4 0 3 0 1] R_4 leftarrow R_4 - 2 R_1 [1 0 0 0 2 0 1 -1 0 0 0 1 2 0 0 0 1] = L [1 0 0 0 -2 -1