Given is
$$A:=\begin{pmatrix} 0 & -2\\ 2 & 0 \end{pmatrix}\in M_{2x2}(\mathbb{R})$$
and let $f:=L(A):\mathbb{R}^2\longrightarrow\mathbb{R}^2$ be the linear map
$1.$ Which geometric interpretation has $f$
$2.$ Find the matrix $M_B, _B(f)$ of $f$ in relation to the basis $B:=(e_2,e_1+e_2)$ in $\mathbb{R}^2$
$3.$ Is there a $n \in \mathbb{N}$ with $\det(A^n)=2017$? Explain your choice.
My attempt:
$1.$ $Ae_1=\begin{pmatrix} 0 \\ 2 \end{pmatrix}$,$\,\, Ae_2=\begin{pmatrix} -2 \\ 0 \end{pmatrix}$
So the map $f$ rotates the input by $90°$ and scales by the factor $2$
$2.$ Im not sure with the notations, but I think what they wanted is:
$_Eid_B=\begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}\Longrightarrow _Bid_E=\begin{pmatrix} -1 & 1\\ 1 & 0 \end{pmatrix}$
$_B(f)=_Bid_E {_EA_E}_Eid_B=\begin{pmatrix} -1 & 1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -2\\ 2 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 2 & 2\\ 0 & -2 \end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 2 & 4\\ -2 & -2 \end{pmatrix}$
$3.$ No there is no $n \in \mathbb{N}$ with $\det(A^n)=2017$
Since $\det(A^n)=\det(A)^n=4^n$ which is diviseable by $4$, $\,\,\,\forall n\in \mathbb{N}$ And $2017$ is not even.
This task was in a older linear algebra 1 exam of a (for me) unknown college. I just wanted to solve it as far as I could, so it would be great if someone could look over it. Actually the notations of the second task are really unknown to me aswell, so I tried making sense of it
Your answers to parts 1 and 3 look good. In part 2, you've misunderstood what you are asked to find. You should find $$ \pmatrix{0&1\\ 1&1}^{-1}A\pmatrix{0&1\\ 1&1}. $$