Linear Algebra True / False with justifications. Which of the following are correct or incorrect?

A) The determinant function $\det:M_n\rightarrow\Bbb R$ is a linear transformation.
False - determinants tell us areas or volumes.
B) If $A$ is an $n \times n$ and the matrix $B$ is obtained from $A$ by switching two rows, then $\det B=(-1)^n\det A$.
False  $\det -A=(-1)^n\det A$  However, this is true if $B=-A$.
C) If $A$ is an $n \times k$ matrix then $\operatorname{rank} A+\operatorname{null}(A)=n$.
False - $\operatorname{rank} A+\dim(\operatorname{null}(A))=n$.
D) The subset $S$ of $P_2$ consisting of all polynomial of the form $p(x)=x^2+14$ is a subspace of $P_2$.
False - Doesn't contain the zero vector.
E) If $A$ is an $n \times n$ matrix, then $\det A^T=\det A$.
True
 A: The one major problem I see is in your reasoning for $(B)$. 
It is indeed false, but it is false because whenever you take $A$ and "swap one row with another" to get $B$, then $\det B = - \det A$, and so for even $n,\;$ the option $\;B = (-1)^n \det A = \det A \neq -\det A$, is indeed false.

Remarks:
I suspect that your justification for (C), which is correct, is likely what was intended by the statement, but your justification makes your answer clear, with respect to why you are concluding that, as literally stated, it is "false".
Suggestion: you need more justification for $(A)$. "False" is correct, but what does the fact that the determinant evaluates to a scalar (which in 2D, its absolute value represents area, and in 3D volume, but what about n-dimensions?) say about the $\det$ function being a linear transformation? You need to provide a justification which explains why it is not a linear transformation.
A: For the record:
(A) holds if and only if $n=1$. (When $n\neq1$, $\det(2I)=2^n\neq2=2\det(I)$ shows it is wrong.)
(B) holds if and only if $n$ is odd or $\det(A)=0$.
(C) is a nonsensical statement as printed, since $\operatorname{null}(A)$ is a subspace, which cannot be added to a number. However taking the dimension of that subspace instead does not make the statement valid, since rank-nullity states $\operatorname{rank} A+\dim(\operatorname{null}(A))=k$ in this case (the dimension of the space of vectors $A$ can be applied to); the statement would only become true for $n=k$.
In the remaining cases the answer by OP is correct.
A: *

*For (A), it is false, but you have to justify why it is not linear. Just because $\det$ says something about areas or volumes doesn't mean it isn't linear, necessarily. In particular, you know that if a function $f$ is linear then $f(A + B) = f(A) + f(B)$. Does that hold for $\det$?

*For (B), the same concern applies. Just because $\det(-A) = (-1)^n \det(A)$, why does that mean that $\det(B() = (-1)^n \det(A)$ is false? It is false, for the record, but you have to justify it more explicitly.

*For (C), firstly, I think that by $N(A)$, they do mean $\dim{\mathrm{nul}(A)}$. However, I would check your answer. What does the rank nullity theorem say about a matrix of size $s \times r$ (in other words, don't be too attached to the letter $n$).

*For (E), you're right, but what is the justification?
