Bloch sphere

Knowing that the bit flip gate X takes |0> to |1>, which of the below could be a valid matrix representation for the bit flip gate?

$A) \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix},\quad B) \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix},\quad C) \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},\quad D) \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$

The answer is D):

$X=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$

Indeed:

$\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}= \begin{pmatrix} 0\\ 1 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix}= \begin{pmatrix} 1\\ 0 \end{pmatrix}$

Given that H (the Hadamard gate) acting on |0> produces |+>, i.e. H|0>=|+>, and that H|1> = |->, which of the following could be a valid matrix representation for the Hadamard gate?

$A)\; \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix},\quad B)\; \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1\\ -1 & 1 \end{pmatrix},\quad C)\; \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix},\quad D)\; \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ -1 & -1 \end{pmatrix}$

The answer is C):

$H=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$

Indeed:

$H=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \end{pmatrix}, \quad \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix}= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ -1 \end{pmatrix}$

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Bloch sphere