Thanks in advance for your help. I have to do a task on my office computer which means I can't install any additional software, so I think a batch file might be the easiest way to do this even if it's not the most efficient (although I welcome other solutions that I might be able to do without installing software on my office computer). Also, please keep in mind that I have very little programming experience in your answers :)

Basically, I have a folder with a series of subfolders in it, each of these full of files of various types (text, emails, text-based pdfs, word docs). I also have a text file (or an excel column) with hundreds of email addresses in it. I want to use a batch file (or similar method) to run through each email address in the list and search against the contents of the various files in each folder like Windows Search does. Files which contain the search phrase should automatically be moved to a prespecified folder (possibly having to be renamed or put in a new subfolder as many of these files might have the same name). In the end, we should have one folder (as well as its subfolders) which contains no files which have any of the email addresses from the list in their contents, and another folder which only has files that has content containing the email addresses in their files (these can be in subfolders, or renamed, or whatever is most convenient, just not deleted).

(Since the list of email addresses is in Excel, another option I could entertain would be using some sort of VBA macro if that is capable of searching the contents of files within a folder recursively.)

I hope that this makes sense and I'm happy to clarify if it doesn't. Again, thanks for any help you can provide!


Question: A Question about Matrix Elementary Row Operation = Scalar multiplication to a row
Answer: Yes, your thought is correct, as specified, because $\,k\neq 0$. 

As you say: multiplying a row by a non-zero scalar $k$ is precisely equivalent to dividing a row by the reciprocal of $k\neq 0,\;$ i.e., dividing the row by $\;\dfrac{1}{k}.$