Is there a way to "fill out" a $n\times n$ matrix so that the determinant is the same as an$ m\times m$ matrix with $m < n$?

Question

Say I have a $3\times3$ matrix: \begin{bmatrix}x&y&1\\0&1&1\\1&1&1\end{bmatrix} The determinant is then $y-1$, is there a way to fill out a strictly larger matrix that still contains the original so that the determinant remains $y-1$?

This is an example of what I am describing but I am not sure that this reflection into the lower left "quadrant" thing I did is true for all matrices.

http://www.wolframalpha.com/input/?i=determinant%5B%7Bx%5E2,xy,y%5E2,x,y,1%7D,%7B0,0,0,0,1,1%7D,%7B0,0,0,1,1,1%7D,%7B0,0,1,0,0,0%7D,%7B0,1,1,0,0,0%7D,%7B1,1,1,0,0,0%7D%5D+%3D+determinant%5B%7Bx,y,1%7D,%7B0,1,1%7D,%7B1,1,1%7D%5D

Answer 1

For example

$$\det\begin{pmatrix}x&y&1\\0&1&1\\1&1&1\end{pmatrix}=\det\begin{pmatrix}x&y&1&0\\0&1&1&0\\1&1&1&0\\0&0&0&1\end{pmatrix}=\det\begin{pmatrix}x&y&1&0&0\\0&1&1&0&0\\1&1&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{pmatrix}=etc.$$