Answer
The third column of $A^{-1}$ is $\left[\begin{array}{l}
3\\
-6\\
4
\end{array}\right]$
Work Step by Step
Row reduce $[A \mathrm{e}_{3}]:$
$\left[\begin{array}{llll}
-2 & -7 & -9 & 0\\
2 & 5 & 6 & 0\\
1 & 3 & 4 & 1
\end{array}\right]\left\{\begin{array}{l}
\leftrightarrow R_{3}.\\
+R_{1}.\\
.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 3 & 4 & 1\\
0 & -2 & -3 & 0\\
-2 & -7 & -9 & 0
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
+2R_{1}.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 3 & 4 & 1\\
0 & -2 & -3 & 0\\
0 & -1 & -1 & 2
\end{array}\right]\left\{\begin{array}{l}
.\\
-3R_{3}.\\
.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 3 & 4 & 1\\
0 & 1 & 0 & -6\\
0 & -1 & -1 & 2
\end{array}\right]\left\{\begin{array}{l}
-3R_{2}.\\
.\\
+R_{2}.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 4 & 19\\
0 & 1 & 0 & 6\\
0 & 0 & -1 & -4
\end{array}\right]\left\{\begin{array}{l}
+4R_{3}.\\
.\\
\times(-1).
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llll}
1 & 0 & 0 & 3\\
0 & 1 & 0 & 6\\
0 & 0 & 1 & 4
\end{array}\right]$ = $[I\ e_{3}^{\prime}]$
The third column of $A^{-1}$ is $\left[\begin{array}{l}
3\\
-6\\
4
\end{array}\right]$
