Answer
No, due to theorems 5 and 8.
Work Step by Step
Th.5. states:
If $A$ is an invertible $n\times n$ matrix, then for each $\mathrm{b}$ in $\mathbb{R}^{n}$, the equation $A\mathrm{x}=\mathrm{b}$ has the unique solution $\mathrm{x}=A^{-1}\mathrm{b}$.
It is in form $p\rightarrow q$. The contrapositive (which is equivalent to it ) is $(\neg q)\rightarrow(\neg p)$
So, if it is not true that $G\mathrm{x}=\mathrm{y}$ has exactly one solution for every y, then G is not invertible.
Then, using the IMT (Th.8),
$\mathrm{a}. \quad A$ is an invertible matrix
and
$\mathrm{h}.\ \quad $The columns of $A$ span $\mathbb{R}^{n}$.
are both false (when A=G).
