Answer
$A$ is not invertible.
Work Step by Step
ALGORITHM FOR FINDING $A^{-1}$
Row reduce the augmented matrix $[A\ I]$.
If $A$ is row equivalent to $I$, then $[A\ I]$ is row equivalent to $[I\ A^{-1}]$.
Otherwise, $A$ does not have an inverse.
---
$\left[\begin{array}{llllll}
1 & -2 & 1 & 1 & 0 & 0\\
4 & -7 & 3 & 0 & 1 & 0\\
-2 & 6 & 4 & 0 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
-4R_{1}.\\
+2R_{1}.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llllll}
1 & 2 & -1 & 1 & 0 & 0\\
0 & 1 & -1 & 4 & 1 & 0\\
0 & -2 & 2 & 2 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
-2R_{2}.\\
.\\
+2R_{2}.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llllll}
1 & 2 & -1 & 1 & 0 & 0\\
0 & 1 & -1 & 4 & 1 & 0\\
0 & 0 & 0 & 10 & 2 & 1
\end{array}\right]$
The bottom row on the LHS has all three zero entries,
so A is not row equivalent to I,
$A$ is not invertible.
