Answer
$6$
Work Step by Step
We find the determinant by performing a series of row exchanges, plus a single row swap. Hence, we reverse the sign of the determinant in deriving the echelon matrix, so we multiply by $-1$ in our computations to compensate.
$\begin{vmatrix}1&3&-1&0&-2\\0&2&-4&-2&-6\\-2&-6&2&3&10\\1&5&-6&2&-3\\0&2&-4&5&9\end{vmatrix}=\begin{vmatrix}1&3&-1&0&-2\\0&2&-4&-2&-6\\0&0&0&3&6\\0&2&-5&2&-1\\0&2&-4&5&9\end{vmatrix}=\begin{vmatrix}1&3&-1&0&-2\\0&2&-4&-2&-6\\0&0&0&3&6\\0&0&-1&4&5\\0&0&0&7&15\end{vmatrix}=-\begin{vmatrix}1&3&-1&0&-2\\0&2&-4&-2&-6\\0&0&-1&4&5\\0&0&0&3&6\\0&0&0&7&15\end{vmatrix}=-\begin{vmatrix}1&3&-1&0&-2\\0&2&-4&-2&-6\\0&0&-1&4&5\\0&0&0&3&6\\0&0&0&0&1\end{vmatrix}=-[1\cdot 2\cdot (-1)\cdot 3\cdot 1]=6$
