Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
5 & 9 & 17 \\7 & 21 & 13\\27 & 16 & 8
\end{bmatrix}$
To find the inverse of $A$, we obtain augmented matrix:
$\begin{bmatrix}
5 & 9 & 17 | 1&0 & 0\\7 & 21 & 13| 0 & 1 & 0 \\ 27 & 16 & 8 | 0 & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & \frac{9}{5}& \frac{17}{5} | \frac{1}{5}&0 & 0\\7 & 21 & 13| 0 & 1 & 0 \\ 27 & 16 & 8 | 0 & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & \frac{9}{5}& \frac{17}{5} | \frac{1}{5}&0 & 0\\0& \frac{42}{5} & -\frac{54}{5}| -\frac{7}{5} & 1 & 0 \\ 0 & -\frac{163}{5} & -\frac{419}{5} | -\frac{27}{5} & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & \frac{9}{5}& \frac{17}{5} | \frac{1}{5}&0 & 0\\0&1& -\frac{9}{7}| -\frac{1}{6} & \frac{5}{42} & 0 \\ 0 & -\frac{163}{5} & -\frac{419}{5} | -\frac{27}{5} & 0 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0& \frac{40}{7} | \frac{1}{2}&-\frac{3}{14} & 0\\0&1& -\frac{9}{7}| -\frac{1}{6} & \frac{5}{42} & 0 \\ 0 & 0 & -\frac{880}{7} | -\frac{65}{6} & \frac{163}{42}& 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0& \frac{40}{7} | \frac{1}{2}&-\frac{3}{14} & 0\\0&1& -\frac{9}{7}| -\frac{1}{6} & \frac{5}{42} & 0 \\ 0 & 0 & 1 | \frac{91}{1056} & -\frac{163}{5280}& -\frac{7}{880}
\end{bmatrix} \approx \begin{bmatrix}
1 & 0&0 | \frac{1}{132}&-\frac{5}{132} & \frac{1}{22}\\0&1&0| -\frac{59}{1056} & \frac{419}{5280} & -\frac{9}{880} \\ 0 & 0 & 1 | \frac{91}{1056} & -\frac{163}{5280}& -\frac{7}{880}
\end{bmatrix}$
Hence, $A^{-1}= \begin{bmatrix}
\frac{1 }{132} & \frac{-5}{132}&\frac{1}{22}\\-\frac{59}{1056} & \frac{419}{5280} & \frac{-9}{880}\\\frac{91}{1056} & -\frac{163}{5280}& -\frac{7}{880}
\end{bmatrix}$
where $\Delta \ne0$
