Answer
$A^{-1}=\left[\begin{array}{rrrrrr}
8 & 3 & 1\\
10 & 4 & 1\\
7/2 & 3/2 & 1/2
\end{array}\right]$
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.
---
$\left[\begin{array}{llllll}
1 & 0 & -2 & 1 & 0 & 0\\
-3 & 1 & 4 & 0 & 1 & 0\\
2 & -3 & 4 & 0 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
+3R_{1}.\\
-2R_{1}.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llllll}
1 & 0 & -2 & 1 & 0 & 0\\
0 & 1 & -2 & 3 & 1 & 0\\
0 & -3 & 8 & -2 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
+3R_{2}.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llllll}
1 & 0 & -2 & 1 & 0 & 0\\
0 & 1 & -2 & 3 & 1 & 0\\
0 & 0 & 2 & 7 & 3 & 1
\end{array}\right]\left\{\begin{array}{l}
+R_{3}.\\
+R_{3}.\\
\div 2.
\end{array}\right.\rightarrow$
$\rightarrow\left[\begin{array}{llllll}
1 & 0 & 0 & 8 & 3 & 1\\
0 & 1 & 0 & 10 & 4 & 1\\
0 & 0 & 1 & 7/2 & 3/2 & 1/2
\end{array}\right]=[I\ A^{-1}]$.
$A^{-1}=\left[\begin{array}{rrrrrr}
8 & 3 & 1\\
10 & 4 & 1\\
7/2 & 3/2 & 1/2
\end{array}\right]$
