I would like some clarification on the lu(A) function.
A = [1, -1, 2; 3, -1, 7; 2, -4, 5] which is invertible and LU-factorable without row interchange, so the LU matrices should be unique.
By hand, I get L = [1, 0, 0; 3, 1, 0; 2, -1, 1] and U = [1, -1, 2; 0, 2, 1; 0, 0, 2], but why do I get the following results? First, the L isn't lower triangular, and second one there is a permutation matrix while A can be made row-echelon without row interchange? What can't I get A = L*U with L in the 'right' format, instead of P*A = L*U.
A =
1 -1 2
3 -1 7
2 -4 5
>> [l, u] = lu(A)
l =
0.33333 0.20000 1.00000
1.00000 0.00000 0.00000
0.66667 1.00000 0.00000
u =
3.00000 -1.00000 7.00000
0.00000 -3.33333 0.33333
0.00000 0.00000 -0.40000
>> [l, u, p] = lu(A)
l =
1.00000 0.00000 0.00000
0.66667 1.00000 0.00000
0.33333 0.20000 1.00000
u =
3.00000 -1.00000 7.00000
0.00000 -3.33333 0.33333
0.00000 0.00000 -0.40000
p =
Permutation Matrix
0 1 0
0 0 1
1 0 0
A = [1, -1, 2; 3, -1, 7; 2, -4, 5] which is invertible and LU-factorable without row interchange, so the LU matrices should be unique.
By hand, I get L = [1, 0, 0; 3, 1, 0; 2, -1, 1] and U = [1, -1, 2; 0, 2, 1; 0, 0, 2], but why do I get the following results? First, the L isn't lower triangular, and second one there is a permutation matrix while A can be made row-echelon without row interchange? What can't I get A = L*U with L in the 'right' format, instead of P*A = L*U.
A =
1 -1 2
3 -1 7
2 -4 5
>> [l, u] = lu(A)
l =
0.33333 0.20000 1.00000
1.00000 0.00000 0.00000
0.66667 1.00000 0.00000
u =
3.00000 -1.00000 7.00000
0.00000 -3.33333 0.33333
0.00000 0.00000 -0.40000
>> [l, u, p] = lu(A)
l =
1.00000 0.00000 0.00000
0.66667 1.00000 0.00000
0.33333 0.20000 1.00000
u =
3.00000 -1.00000 7.00000
0.00000 -3.33333 0.33333
0.00000 0.00000 -0.40000
p =
Permutation Matrix
0 1 0
0 0 1
1 0 0