I'm trying to get ready to my exam from linear algebra by doing some random tasks and with this one i'm pretty stuck.
$$A = \begin{pmatrix} 8-5i & -6 \\ 4-5i & -3+5i \end{pmatrix}$$
Given this matrix $A$ i have to calculate determinant of $A^{-2013}$. I tried to do this by using eigenvalues. And i know this matrix has inversion ( since $det(A) \neq 0 $ ), so it's true that $det (A^{-2013}) = det (A^{2013})^{-1}$. What more i got is that eigenvalues are $\lambda_1 = 5-5i$ and $\lambda_2 = 5i$. With that, we can use similarity matrix B that
$$B= \begin {pmatrix} 5-5i & 0 \\ 0 & 5i \end{pmatrix} $$ and $$A = P^{-1}\times B \times P$$ so we have to find $A \times C = B$ And this is the part i'm stuck with…
I have no idea how to do it nicely, because by staight matrix multiplication i end up with 4 complicated equations. From that i think it's okay to do $ ( P^{-1} \times B \times P ) ^{2013} = (P^{-1} \times B \times P \times P^{-1} \times B \dots ) = P^{-1} \times B^{2013} \times P $, so the final answer will be $det(P^{-1} \times B^{2013} \times P)^{-1}$, since $B$ is diagonal matrix, raising it to 2013th power will be easy. So, please help me figuring out this hard part, i don't want to make it overcomplicated 🙂 Thanks in advance!!!
Best Answer
This is much simpler, $\det(A^n)=\det(A)^n$. And consider also that $(1-i)^2=-2i$.