In his April 26 submission "More Matrix Results on HP and TI Machines" Valentin Albillo proposed a seventh order matrix to be known as "Matrix No. 1" as a better test of computation capability than a seventh order Hilbert. In another April 26 submssion "There you are" Valentin proposed two more seventh order matrices, "Matrix No. 2" and "Matrix No. 3". In an April 27 submission "(LONG) Commented Results for HP-71B (and HP-15C)" he published the exact inverse for his No. 1 matrix. Exact inverses for matrices No. 2 and No. 3 have not been published to date. The exact inverses for those matrices as calculated using the exact mode on the HP-49 are:

Exact Inverse for Albillo's Matrix No. 2 :193969587 693099 -1665228 3111895 -14779418 67861431 -194191637

7798 28 -67 125 -594 2728 -7807

-64638 -231 555 -1037 4925 -22614 64712

504013 1801 -4327 8086 -38403 176332 -504590

-4282933 -15304 36769 -68712 326336 -1498410 4287836

25738861 91971 -220968 412934 -1961160 9004896 -25768326

-193970402 -693102 1665235 -3111908 14779480 -67861716 194192453

Exact Inverse for Albillo's Matrix No. 3:294228951 702720 -1942323 4764003 -21337482 76325631 -294398186

12108 29 -80 196 -878 3141 -12115

-90417 -216 597 -1464 6557 -23455 90469

512920 1225 -3386 8305 -37197 133056 -513215

-4325606 -10331 28555 -70038 313693 -1122101 4328094

36887482 88100 -243509 597263 -2675080 9568944 -36908699

-294230620 -702724 1942334 -4764030 21337603 -76326064 294399856

I had previously noticed that the first and last elements for every row and every column of the inverse of Matrix No.1 were approximately equal and larger than any other element in the row or column. I was surprised to see the same characteristic in the inverses of Matrix No. 2 and Matrix No. 3. When I looked more closely at the inverses I saw patterns in the individual inverses and among the inverses. The following table shows the results when I divide each element of the seventh row of Matrix No. 1 by the corresponding element of the sixth row, and do the same for the seventh and sixth column.

Element row 7 / row 6 col 7 / col 61 -7.32762314181 -3.19606552693

2 -7.32758186686 -3.19586894587

3 -7.32762522837 -3.19606732481

4 -7.32762448362 -3.19606423358

5 -7.32762198907 -3.19606453553

6 -7.32762262334 -3.19606533695

7 -7.32762303367 -3.19606551596

I found similar consistencies in every set of ratios that I tried. If I do the row 7 divided by row 6 calculations for the inverses of Matrices No. 2 and No. 3 I get ratios of -7.53609... and -7.97643... If I do the column 7 divided by column 6 calculations for the inverses of Matrices No. 2 and No. 3 I get ratios of -2.861... and -3.857... .

When I worked with problems like the one near the end of Kahan's "Mathematics Written in Sand" , namely an 8x8 modified Hilbert where A(i,j) = 360360/(i + j - 1) I recognized that the test matrix was symmetric and expected that the exact inverse would also be symmetric, which it was. I could also see evidence of approximate symmetry in the non-exact inverses as delivered from my HP-28S and TI-59. So, I was surprised to find that I could see patterns in the inverses of the Albillo matrices while I could not see patterns in the original matrices that Valentin had described as random. .

In the thread that introduced the Albillo matrices Werner asked twice for information on how the matrices were formed. I, too, would like to have more information on how these matrices were formed.