# Solution 4.2 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

## Education Article

**Published:**Jan 1, 2000**Channels:**Chemometrics & Informatics

1. The centred data matrix is given below.

-0.500 |
-0.857 |
-0.300 |
-3.400 |
-2.557 |
-0.657 |

-0.600 |
-1.457 |
1.600 |
3.400 |
0.243 |
-0.257 |

1.100 |
2.943 |
-1.800 |
0.000 |
1.243 |
0.343 |

-0.700 |
-1.657 |
0.500 |
-0.300 |
-1.557 |
-0.557 |

0.800 |
1.043 |
-0.600 |
-2.000 |
0.243 |
1.143 |

-0.100 |
0.143 |
0.300 |
2.700 |
1.743 |
0.343 |

0.000 |
-0.157 |
0.300 |
-0.400 |
0.643 |
-0.357 |

The loadings can be calculated by ** P** = (

**)**

*T'.T*^{-1}.

**.**

*T'***and are as follows**

*X*
-0.0318 |
-0.0449 |
0.1988 |
0.8901 |
0.4054 |
0.0272 |

0.3054 |
0.7080 |
-0.3848 |
-0.0803 |
0.4532 |
0.2133 |

2. The scores vectors are orthogonal because the sum of the product of the elements of the two scores vectors equals 0. The same property is exhibited for the loadings vectors. In addition the sum of squares of the elements of each loading vector equals 1. Note that there may be some small round errors.

3. The eigenvalues can be obtained by determining the sum of squares of the scores for each PC, namely 43.088 and 30.051 respectively. The overall sum of squares for the centred data is 76.251. Note that it is important to use the centred rather than raw data. The percentage variance of the first two PCs is given by 100 ´ (43.088 + 30.051) / 76.251 or 95.917%.