Given
A = b =
| 1 2 4 8 | | 10 |
| 1/2 3 6 12 | | 15 |
| 1/4 1 2 12 | | 13 |
The LU factorization of A is
L =
| 1 0 0 |
| 1/2 1 0 |
| 1/4 1/4 1 |
U =
| 1 2 4 8 |
| 0 2 4 8 |
| 0 0 0 8 |
Reduced echelon of A is :
| 1 0 0 0 |
| 0 1 2 0 |
| 0 0 0 1 |
Amount of independant collumns :
i think 2 -- collumn 2 and 3 are dependant
The basis for the zerospace :
x1 = 0
x2 = -2x3
x4 = 0
thus
| 0 |
| -2 |
| 1 | (free)
| 0 |
The bases for the column space is :
| 1 2 8 |
| 1/2 3 12 |
| 1/4 1 12 |
Asked :
*Vector b is the sum of the 2de and 4the column of A. What is the solution set of Ax=b? Without using ANY EXTRA calculations. Why do i have all the ingredients for writing the solution immediatley ? *
Answer from matlab:
| 0 |
| 0 |
| 1/2 |
| 1 |
Since $b = a_2 + a_4$ and $2a_2 = a_3$ you have $$A(e_2+e_4) = A(\frac{1}{2}e_3 + e_4) = b$$ Therefor the Solution set is the 1-dimensional affine subspace containing $$e_2+e_4, \frac{1}{2}e_3+e_4$$