Solution 6.4 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant
Education Article
- Published: Jan 1, 2000
- Channels: Chemometrics & Informatics
1. The following graph is obtained
2. The following data should be obtained.
The graph is as follows.
3. The graph is presented below
The values close to 0 are regions of least change in the spectrum and so regions of most purity. The points 6, 17 and 26 are minima and so represent the centre of regions of least change and highest purity.
4. The graph of the resolved profiles using MLR is as follows.
5. The eigenvalues are given below, together with the contribution to percentage sum of square errors.
PC |
E-value |
% |
1 |
84.906 |
89.274 |
2 |
5.588 |
5.876 |
3 |
4.346 |
4.570 |
4 |
0.0916 |
0.096 |
5 |
0.0627 |
0.066 |
There is a sharp dip in the size of the eigenvalue after the third component.
6. Step a is quite easy to do since the matrix P.P' is simply a unit matrix. Simply calculate R^{-1} = S P', to give the inverse the R matrix as follows
1.792 |
0.842 |
-0.300 |
3.126 |
-0.650 |
-0.300 |
1.314 |
0.208 |
0.807 |
so that R is given by
0.124 |
0.199 |
0.120 |
0.781 |
-0.492 |
0.107 |
-0.402 |
-0.197 |
1.016 |
The three profiles are presented below. The appear slightly cleaner to the profiles obtained using MLR, but there is not much difference.