The correct answer is $\boxed{\frac{1}{3}\left[ {\begin{array}{*{20}{c}} 7&2 \ -5&1 \end{array}} \right]}$.
To find the inverse of a 2×2 matrix, we can use the formula $A^{-1} = \frac{1}{|A|}\left[ {\begin{array}{*{20}{c}} \left| \begin{array}{cc} a_{2,2} & a_{2,1} \\ a_{1,2} & a_{1,1} \end{array} \right| & \left| \begin{array}{cc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{array} \right| \\ \left| \begin{array}{cc} a_{2,2} & a_{2,3} \\ a_{1,2} & a_{1,3} \end{array} \right| & \left| \begin{array}{cc} a_{1,3} & a_{1,2} \\ a_{2,3} & a_{2,2} \end{array} \right| \\ \end{array}} \right]$, where $a_{i,j}$ is the element of $A$ in the $i$th row and $j$th column.
In this case, $A = \left[ {\begin{array}{{20}{c}} 1&2 \ 5&7 \end{array}} \right]$, so $|A| = \left| \begin{array}{cc} 7 & 2 \\ 5 & 1 \end{array} \right| = 3$. Therefore, the inverse of $A$ is $\frac{1}{3}\left[ {\begin{array}{{20}{c}} 7&2 \ -5&1 \end{array}} \right]$.
Here is a step-by-step solution:
- Find $|A|$.
$|A| = \left| \begin{array}{cc} 7 & 2 \\ 5 & 1 \end{array} \right| = 3$
- Divide $|A|$ by itself to find $A^{-1}$.
$A^{-1} = \frac{1}{|A|}\left[ {\begin{array}{{20}{c}} 7&2 \ -5&1 \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{{20}{c}} 7&2 \ -5&1 \end{array}} \right]$