Answer
$-3$
Work Step by Step
$\begin{vmatrix}1&5&-4\\-1&-4&5\\-2&-8&7\end{vmatrix}=\begin{vmatrix}1&5&-4\\0&1&1\\0&2&-1\end{vmatrix}=\begin{vmatrix}1&5&-4\\0&1&1\\0&0&-3\end{vmatrix}=1\cdot 1\cdot (-3)=-3$
Note that we only added constant multiples of some rows to others, so the determinant of the echelon matrix equals the determinant of the original matrix.
