Answer
$A$ is not an invertible matrix.
(Th.8, (a) and (b))
Work Step by Step
Swapping rows 1 and 2,
$A\sim\left[\begin{array}{lll}
1 & 0 & 2\\
0 & 3 & -5\\
-4 & -9 & 7
\end{array}\right] $, adding $4R_{1}$ to $R_{3}$
$\sim\left[\begin{array}{lll}
1 & 0 & 2\\
0 & 3 & -5\\
0 & -9 & 15
\end{array}\right]$ adding $3R_{2}$ to $R_{3}$
$\sim\left[\begin{array}{lll}
1 & 0 & 2\\
0 & 3 & 5\\
0 & 0 & 0
\end{array}\right]$,
we see that $A$ can not row-reduce to $I_{3}$
So, looking at Th.8,
$\mathrm{b}. \quad A$ is row equivalent to the $n\times n$ identity matrix..
is not valid. Then,
$\mathrm{a}. \quad A$ is an invertible matrix.
is not valid as well
$A$ is not an invertible matrix.
(Th.8, (a) and (b))