To use the Cayley-Hamilton theorem, we first compute the characteristic polynomial $p(t)$ of the matrix $A$. We have
\begin{align*}
&p(t)=\det(A-tI)\\
&\begin{vmatrix}
1-t & 1 & 2 \\
9 &2-t &0 \\
5 & 0 & 3-t
\end{vmatrix}\\[6pt]
&=(-1)^{3+1}5\begin{vmatrix}
1 & 2\\
2-t& 0
\end{vmatrix}+(-1)^{3+2}\cdot 0 \begin{vmatrix}
1-t & 2\\
9& 0
\end{vmatrix}+(-1)^{3+3}(3-t)\begin{vmatrix}
1-t & 1\\
9& 2-t
\end{vmatrix} \\[6pt]
& \text{(by the 3rd row cofactor expansion)}\\
&=5(2t-4)+0+(3-t)\left(\, (1-t)(2-t)-9 \,\right)\\
&=-t^3+6t^2+8t-41.
\end{align*}
Then the Cayley-Hamilton theorem yields that $p(A)=O$, the zero matrix. That is, we have
\begin{align*}
O=p(A)=-A^3+6A^2+8A-41I.
\end{align*}
Thus, we have
\[41I=-A^3+6A^2+8A=A(-A^2+6A+8I),\]
or equivalently
\[I=A\left(\, \frac{1}{41}(-A^2+6A+8I) \,\right).\]
It follows that the inverse matrix is given by
\[A^{-1}=\frac{1}{41}(-A^2+6A+8I).\]
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