1. In the name of GOD

2. Some about model-based Analysis 6 th Iranian Chemometrics workshop Institute for Advances studies in Basic Sciences (IASBS), Zajan, Iran Feb 2007 By: Mohsen Kompany-Zareh

3. Three optically active components in a sample, absorbance spectrum in three wavelengths a 1 = c 1 s 11 + c 2 s 12 + c 3 s 13 a 2 = c 1 s 21 + c 2 s 22 + c 3 s 23 a 3 = c 1 s 31 + c 2 s 32 + c 3 s 33 or a = S c S -1 a = S -1 S cSolution: Square Non-singular S -1 a = c S -1 S=I

4. a = S c What if S is not square? S -1 S=I (for square nonsing. S) S -1 a = S -1 S c ?S=I (for not square S) S T S is square  (S T S) -1 S T S =I (S T S) -1 S T pseudo inverse S + S=I S + a = c S -1 a = c

5. S + a = c a = S c = S S + a Projection of a in space of S  a a r = || - a||  0 if a is in space of S

6. X = C S C = X Z X C classical inverse To fit the parameters that form C = X X + C = C C + X

7. C r X r = || - C|| = 0.08 C= f(K) K=2 = X X + C

8. C r X r = || - C|| = 0.04 C= f(K) K=3 = X X + C

9. C r X r = || - C|| = 0.01 C= f(K) K=4 = X X + C

10. C r X r = || - C|| = 0.0001 C= f(K) K=5 = X X + C

11. X r C r = || - X|| = 0.15 C= f(K) K=2 = C C + X

12. X r C r = || - X|| = 0.08 C= f(K) K=3 = C C + X

13. X r C r = || - X|| = 0.011 C= f(K) K=4 = C C + X

14. X r C r = || - X|| = 0.001 C= f(K) K=5 = C C + X

15. M +LML [M] [L] [ML K f = [M] [L] [ML] K f C L = [L] + [ML] C M = [M] + [ML] One-step complex formation equilibrum CL = [L] + Kf [M] [L] CM = [M] + Kf [M] [L] K f [L] 2 + (K f C M -K f C L +1)[L] – C L =0 Estimation of [L] at any K f, C M and C L

16. M L Spectroph.

17. = X X + [L] r = || - [L]||  0

27. log(K f )=4