# 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

Staff Emeritus
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

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