MHB What is the General Pattern for Finding Matrix Powers?

[Aryth1]

I'm sure that this problem is easier than I am making out to be, but I'm going over some review problems for an exit exam and I'm having a little trouble with this one.

Let the matrix $A$ be given by:

$$A = \begin{pmatrix} 1&4\\ 2&3 \end{pmatrix}$$

Find $A^n$ for general $n$.

I have the first three iterations done, but I've been unable to find the pattern.

$$A^2 = \begin{pmatrix} 1&4\\ 2&3 \end{pmatrix}\begin{pmatrix} 1&4\\ 2&3 \end{pmatrix} = \begin{pmatrix} 9&16\\ 8&17 \end{pmatrix}$$

$$A^3 = \begin{pmatrix} 9&16\\ 8&17 \end{pmatrix}\begin{pmatrix} 1&4\\ 2&3 \end{pmatrix} = \begin{pmatrix} 41&84\\ 42&83 \end{pmatrix}$$

$$A^4 = \begin{pmatrix} 41&84\\ 42&83 \end{pmatrix}\begin{pmatrix} 1&4\\ 2&3 \end{pmatrix} = \begin{pmatrix} 209&416\\ 208&417 \end{pmatrix}$$

Any help in finding the general pattern would be greatly appreciated!

 

[I like Serena]

I'm sure that this problem is easier than I am making out to be, but I'm going over some review problems for an exit exam and I'm having a little trouble with this one.

Let the matrix $A$ be given by:

$$A = \begin{pmatrix} 1&4\\ 2&3 \end{pmatrix}$$

Find $A^n$ for general $n$.

I have the first three iterations done, but I've been unable to find the pattern.

$$A^2 = \begin{pmatrix} 1&4\\ 2&3 \end{pmatrix}\begin{pmatrix} 1&4\\ 2&3 \end{pmatrix} = \begin{pmatrix} 9&16\\ 8&17 \end{pmatrix}$$

$$A^3 = \begin{pmatrix} 9&16\\ 8&17 \end{pmatrix}\begin{pmatrix} 1&4\\ 2&3 \end{pmatrix} = \begin{pmatrix} 41&84\\ 42&83 \end{pmatrix}$$

$$A^4 = \begin{pmatrix} 41&84\\ 42&83 \end{pmatrix}\begin{pmatrix} 1&4\\ 2&3 \end{pmatrix} = \begin{pmatrix} 209&416\\ 208&417 \end{pmatrix}$$

Any help in finding the general pattern would be greatly appreciated!

Hi Aryth! :)

One way to do it is to diagonalize. $A$ can be written as:
$$A=BDB^{-1}$$
where $D$ is a diagonal matrix.

Then we have:
$$A^n = BD^nB^{-1}$$

 

[Aryth1]

Hi Aryth! :)

One way to do it is to diagonalize. $A$ can be written as:
$$A=BDB^{-1}$$
where $D$ is a diagonal matrix.

Then we have:
$$A^n = BD^nB^{-1}$$

I had totally forgotten about diagonalizing the matrix. I got it, thanks for the help!