Answer
A is singular.
Work Step by Step
$1.\quad $Start with $[A|I]$,
$2.\quad $ Row reduce,
$3.\quad $ If the reduced form is $[I|B], $then $B=A^{-1}.$
If not, then A is singular (has no inverse)
----
$\left[\begin{array}{lllll}
1 & 1 & | & 1 & 0\\
6 & 6 & | & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
-6R_{1}.
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 1 & | & 1 & 0\\
0 & 0 & | & -6 & 0
\end{array}\right]$
The bottom row entries on the left are all zeros.
We can not generate a [ 0 1 ... ] with row operations.
The form $[I|B]$ can not be achieved.
A is singular.
