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- %OP%VS4.01 (22-Oct-91), Gerald Fitton, R4000 2915 6625 6368
- %CO:A,72,72%
- %C%Arrays - Part 3
- %C%by Gerald L Fitton
- Keywords:
- Array Matrix Fitton
-
- This is Part 3 of the "Arrays" series. Parts 1, and 2 are in the
- directories Array1, and Array2. You should read them first.
-
- Matrix Multiplication
- As we have seen the array multiplication function multiplies element by
- element. Matrix multiplication is different.
-
- Load the file [Array03]. You will discover that I have 'hidden' two
- matrices is slots B10 and E10 using the technique explained in the
- directory Array1. One difference in this sheet is that the second
- matrix, the one in E10, has five columns and two rows.
-
- Place the caret in slot E20 and you will see that the formula is not
- B10á*áE10 (as in [Array02]) but is m_mult(B10,E10). This is the matrix
- multiplication formula. I have expanded the array of E20 into the
- range E22I26 so that you can see what has happened. Under matrix
- multiplication the contents of part of E20 is shown in E22I26 as:
-
- E22á=á(1*11)á+á(6*12)áááF22á=á(1*13)á+á(6*14)áááG22á=á(1*15)á+á(6*16)
- E23á=á(2*11)á+á(7*12)áááF23á=á(2*13)á+á(7*14)áááG23á=á(2*15)á+á(7*16)
- E24á=á(3*11)á+á(8*12)áááF24á=á(3*13)á+á(8*14)áááG24á=á(3*15)á+á(8*16)
-
- You will see that matrix multiplication is a mixture of multiplication
- and addition; it is certainly not element by element. If you are not
- familiar with matrix multiplication then you might find it difficult to
- see the logic behind it. Perhaps a further example will make it clear.
-
- In the file [Array03a] you will find an example of applying matrix
- multiplication to the problem of assessing two candidates called A and
- B. Each candidate takes five tests numbered 1 to 5 which are marked
- fairly. However, the three adjudicators, a, b and c, on the test body
- cannot agree on the relative importance of the five tests so they
- provide three different 'weighting' schemes. Which candidate is best
- under which scheme? Load [Array03a] to find out.
-
-