Solution 6.4 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

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  • 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.

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