Computing Determinant of a Matrix $\textrm{det}(A^{101} - A)$

Question

Let $A$ be a $n$-by-$n$ matrix with real entries such that $A^{-1} = 3A$. What is the determinant of $A^{101} - A$ i.e. $\textrm {det}(A^{101} - A)$?

My attempt:
$$A^{-1} = 3A \Rightarrow 3A^{2} = I \Rightarrow A^{2} = \frac{1}{3}I$$ \begin{align} \textrm{det}(A^{101} - A) & = \textrm{det}[A(A^{100}-I)] \\ & = \textrm{det}(A)\textrm{det}(A^{100}-I) \\ & = \textrm{det}(A)\textrm{det}[(A^{2})^{50}-I] \\ & = \textrm{det}(A)\textrm{det}[(\frac{1}{3}I)^{50}-I] \\ & = \textrm{det}(A)\textrm{det}[(\frac{1}{3^{50}})I-I] \\ & = \frac{1}{3^{n}}\textrm{det}(A^{-1})\textrm{det}[(\frac{1}{3^{50}})I-I] \\ \end{align} Any hints on how to continue from here will be appreciated.

Answer 1

Use: $$\det (A^2)=\det(A\cdot A)=\det(A)\cdot \det(A)=\det(\frac13I) \Rightarrow \det(A)=\pm\frac1{3^{n/2}} $$ Hence: $$\textrm{det}(A)\textrm{det}[(\frac{1}{3^{50}})I-I] =\pm\frac1{3^{n/2}}\cdot \left(\frac1{3^{50}}-1\right)^n. $$ Checking with $A_{1\times 1}=\left(-\frac{1}{\sqrt{3}}\right)$: $$A^{101}-A=\left(-\frac{1}{\sqrt{3}}\right)^{101}-\left(-\frac{1}{\sqrt{3}}\right)=\left(-\frac{1}{\sqrt{3}}\right)\left(\frac{1}{3^{50}}-1\right).$$