# For allimom

A possible solution to allimom’s maths problem:

It’s a while since I’ve done matrices, so I may have forgotten some of the symbolic conventions, but I think what you’re being asked to prove is that is true for all values of n.

n=0 and n=1 are trivial, of course.

For n=2, Proof by induction involves showing that an inital case is true (in this case n=2), and then saying if we assume that the equation is true for n, does it follow that it’s true for n+1. If so, then, as we know it’s true for n=2, then it must be true for n+1=3, and that means if it’s true for n=3, it must be true for n+1=4, and so on.

So, assuming the equation is true for n: i.e. if it’s true for n, then it’s true for n+1, so if it’s true for n=2, then it’s true for all n.

As one of my maths lecturers used to say, W5! (“Which Was What Was Wanted”)

I’ve only put in the minimum of working for the matrices – probably you should expand that a bit, depending on what your instructor normally expects.

Of course, I may have totally misinterpreted the question, but at least I had fun stretching my brain this morning trying to dredge matrices and proof by induction out of the depths where they’ve been hiding all these years.

Hope this helps, allimom, and everyone else can now start reading again, the scary maths stuff is over 🙂

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