The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 0&1&0 \ 1&0&0 \ 0&0&1 \end{array}} \right]}$.
The inverse of a 3×3 matrix can be found using the formula $A^{-1} = \frac{1}{|A|} \begin{bmatrix}
a_{2,2}a_{3,3}-a_{2,3}a_{3,2} & a_{1,3}a_{3,2}-a_{1,2}a_{3,3} & a_{1,2}a_{2,3}-a_{1,3}a_{2,2} \\
a_{2,3}a_{3,1}-a_{2,1}a_{3,3} & a_{1,1}a_{3,3}-a_{1,3}a_{3,1} & a_{1,3}a_{2,1}-a_{1,1}a_{2,3} \\
a_{2,1}a_{3,2}-a_{2,2}a_{3,1} & a_{1,2}a_{3,1}-a_{1,1}a_{3,2} & a_{1,1}a_{2,2}-a_{1,2}a_{2,1}
\end{bmatrix}$, where $a_{i,j}$ is the element of $A$ in the $i$th row and $j$th column.
In this case, $A = \left[ {\begin{array}{{20}{c}} 0&1&0 \ 1&0&0 \ 0&0&1 \end{array}} \right]$. Therefore, $|A| = 1$ and $A^{-1} = \left[ {\begin{array}{{20}{c}} 0&1&0 \ 1&0&0 \ 0&0&1 \end{array}} \right]$.