Use the method of Example 6 to compute the matrix A10, where

Question

Use the method of Example 6 to compute the matrix A¹⁰, where

Answer 1

The goal here is going to be to write the matrix as a diagonal matrix, since taking powers of a diagonal matrix is very easy -- you just exponentiate everything on the diagonal. More specifically, we want to write our matrix M as

M = P * D * P^(-1)

where P is a change of basis matrix and D is a diagonal matrix.

If a matrix M has n linearly independent eigenvectors (which this matrix does), then we can diagonalize it and have eigenvalues on the diagonal. It is important, when diagonalizing, to keep track of the operations which allow you to diagonalize, since doing all of these operations to an identity matrix will give you the change-of-basis matrix necessary to write the original matrix as a diagonal matrix (i.e. in the eigenbasis).

In this case, this matrix will become D = ((1,0),(0,2)) when diagonalized (I'm writing the matrix as a list of row vectors)

The matrix which diagonalizes will be P = ((1,0),(7,1)), and its inverse is P^(-1) = ((1,0),(7,1)).

We can then check that we can write our original matrix M as

M = P * D * P^(-1)

From here, we get the equation:

M^(10) = [ P * D * P^(-1) ]^(10)

= P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1) * P * D * P^(-1)

But the internal P's and P^-(1)'s next to each other will cancel out, and so we get that this is equal to

M*(10) = P * D * D * D * D * D * D * D * D * D * D * P^(-1)

= P * D^(10) * P^(-1)

D^(10) should be pretty easy to compute, since all you need to do is exponentiate the diagonal entries by 10.

From there, you can quickly compute P * D^(10) * P^(-1)