Permanent as projection of determinant and another permanent
A way to have such an explicit example is the following: Consider a matrix $A$ full of indeterminates (you can specialize the entries if you wish). Then use some formula for the permanent of this matrix (Ryser's or Glynn's for example). From this formula, you can build another matrix $B$, of quite huge dimensions, whose determinant equals the formula, thus the permanent of $A$.

Another thing you can do is to consider a matrix A whose determinant and permanent are the same (adjacency matrix of a graph whose cycles are of odd length). Then you can use different constructions to build a new matrix B with the same determinant. For instance, in our paper Symmetric Determinantal Representation of Formulas and Weakly-Skew Circuits, we give an algorithm that given a matrix $A$ builds a symmetric matrix $B$ such that $\det A=\det B$ (Section 3.2).  

You can have a look at our paper for the classical techniques to do what you want (if I am not mistaken). For the second part, it is easy to change the constructions giving a matrix $B$ whose determinant equals some other determinant or formula, in such a way that you obtain another matrix $C$ such that $\operatorname{per} C=\det B$.

A last word to be more explicit:

$$\operatorname{per}\begin{pmatrix} x&amp;y\\ 1&amp;1\end{pmatrix}
   = \det\begin{pmatrix} x&amp;0&amp;0&amp;1 \\ 0&amp;y&amp;0&amp;1 \\ 0&amp;0&amp;1&amp;0 \\ 1&amp;1&amp;0&amp;0\end{pmatrix}$$

I hope this helps, even though I am not quite sure to exactly understand what you mean by "explicit example". If you are looking for an explicit example where for some $(n\times n)$ matrix $A$, the smallest matrix $B$ such that $\operatorname{per} A=\det B$ has dimensions $&gt;n$, then you should look at Mignon &amp; Ressayre's paper "A quadratic bound for the determinant and permanent problem" or subsequent works by Jin-Yi Cai and coauthors.