One of the benefits of writing a blog is that people write to me and tell me all kinds of interesting things.\u00a0 For example, early this morning a colleague of mine sent me the details of a bug in MATLAB 2009b which you can reproduce as follows.<\/p>\n
A=[ 0 2 ; 1 0 ];\r\nb=[ 2 ; 0 ];\r\nA'\\b\r\n\r\nans =\r\n0\r\n1<\/pre>\nwhich is, quite simply, wrong! The previous version of MATLAB, MATLAB 2009a, (along with MATLAB’s free clone – Octave) gets it right though:<\/p>\n
ans =\r\n0\r\n2<\/pre>\nso this is a new bug. Curiously, there is no problem with other matrices. For example the following, correct, result is given by 2009b:<\/p>\n
A = [1 2 ; 1 1]\r\nA'\\b\r\n\r\nans =\r\n-2\r\n 4<\/pre>\nIn case you don’t speak MATLAB, the command A’\\b solves the matrix equation AT<\/sup>x = b for an unknown vector, x.\u00a0 If you speak Mathematica then the equivalent command is<\/p>\n
LinearSolve[Transpose[A], b]<\/pre>\nNow, the MATLAB command transpose(A)\\b is syntactically equivalent to A’\\b and yet the former gives the correct result in this particular case:<\/p>\n
transpose(A)\\b\r\nans =\r\n 0\r\n 2<\/pre>\nWhat I find interesting about this particular bug is that it involves one of MATLAB’s core areas of functionality – solving linear systems. I wonder what readers of this blog think about it? Is it a trivial bug that is unlikely to cause problems in the real world or is it something that we should be worried about?<\/p>\n