Homework Help: Matrix with power of n

1. Oct 9, 2009

MechaMZ

1. The problem statement, all variables and given/known data
a and n could be any real number, please evaluate the value.

$$B =\left [ \begin{array}{ccc} a&1&0\\ 0&a&0\\ 0&0&a \end{array} \right ]$$n

I've no idea how to approach this question, because i have never came across the matrix with power not equals to -1.
could somebody give me some hints?

I've try to let n = -1 by using gauss jordan elimination, and let then elements become 1/a, 1/a2 and etc. but this seems incorrect and not what the question asking for..

2. Oct 9, 2009

Dick

You can either expand it for small values of n and guess the pattern and prove it works by induction (if you have to). Or write the matrix as aI+J, where I is the identity and J is zero except for that off diagonal 1. Now, I and J commute, so you can apply the binomial theorem to (aI+J)^n.

3. Oct 9, 2009

Staff: Mentor

Are you sure that the problem says that n could be any real number? You're going to have a hard time raising your matrix to the power $\pi$, say.

4. Oct 9, 2009

MechaMZ

yeah, the question states the a and n is any real value. but why power of pi?

but i'm not very sure this question is asking me to find the n and a value or prove it could be done by any real value(for a and n)

5. Oct 9, 2009

Staff: Mentor

I understand that the problem is not asking you to find a and n, but if n can be any real number, then there is nothing that prevents n from being $\pi$. If so, what does it mean to raise a matrix to an irrational power? A more reasonable interpretation, IMO, would be that n is a positive integer, for which math induction would be one approach.

6. Oct 9, 2009

HallsofIvy

I'm with the others. If both a and n can be any real numbers, then for some combinations of a and n, the matrix won't even exist. For example, if a= 0 and n= -1, B-1 will not exist. If a is negative and n = 1/2, you are going to run into complex numbers. If a can be any real number, and n any positive integer, the problem is relatively simple. Calculate B, B2, B3, make a good guess at what the general form will be and, as Mark44 suggested, prove it by induction.

7. Oct 9, 2009

Dick

Boo hoo. Nobody likes my binomial theorem suggestion??

8. Oct 9, 2009

Staff: Mentor

I like it. I was just offering another possible slant.

9. Oct 9, 2009

MechaMZ

yeah, i like this solution =)
thanks all