Let $V,W$ be finite-dimensional $\mathbb{K}$-vector spaces of dimension $n \gt 0$ and $\varphi:V\to W$ a linear map.
a) Prove: for every ordered basis $B$ of $V$ and every ordered basis $C$ of $W$ - $\varphi$ is an isomorph map of vector spaces if $det(M_{C,B}(\varphi)) \neq 0$
b) Assume, $\varphi$ is an isomorph map. Prove that for every $\lambda\in \mathbb{K}$ \ {$0$} you can find ordered basis $B$ and $C$ so that $det(M_{C,B}(\varphi)) =\lambda$
c) prove that $det(\lambda A) = \lambda^{n}det(A)$ is true for every $\lambda \in \mathbb{K},A \in \mathbb{K}^{n\times n}$
(I translated from german so if anything doesn't make sense, let me know)
I would appreciate any hints on how to solve any of these.
Update
My attempt:
a) In our script it says that $\varphi:V\to W$ is bijective and an isomorph map of vector spaces only if the Matrix $M_{B_{W},B_{V}}(\varphi)$ is invertible. And if the determinant is not 0, the matrix is invertible.
b) I'm not sure about that yet
c) My idea is to use the Leibniz formula for determinants.
$det(\lambda A) = \lambda^{n} * det(A)$
proof
$\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} \lambda a_{\sigma(i),i} = \lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$
extracted the $\lambda$ from the multiplication
$\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\lambda^{n} \prod_{i=1}^{n} a_{\sigma(i),i} = \lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$
factored out the $\lambda$
$\lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma) \prod_{i=1}^{n} a_{\sigma(i),i} = \lambda^{n}\sum_{\sigma\in S_{n}}^{} sgn(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$
$a)$ This is fine.
$b)$ Note that if $B = \{v_1,\dots,v_n\}$ is any basis for $V$, then $C = \{\varphi(v_1),\dots,\varphi(v_n)\}$ is a basis for $W$ (since $\varphi$ is an isomorphism) and $\textsf{M}_{C,B}(\varphi)$ is the identity matrix. Using the same idea, you can obtain the $n \times n$ matrix $$ \begin{pmatrix} \lambda & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix}. $$
$c)$ This is also fine, but I suggest not start with the equality $\det(\lambda A) = \lambda^n \det(A)$, and instead just explain the steps (as you did) to arrive at $\lambda^n \det(A)$ from the expression for $\det(\lambda A)$.