Answer
$-48$
Work Step by Step
We solve by performing elementary row operations on the matrix until the original $4\times 4$ determinant reduces to a simple $2\times 2$ determinant.
$\begin{vmatrix}3&4&-3&-1\\3&0&1&-3\\-6&0&-4&3\\6&8&-4&-1\end{vmatrix}=\begin{vmatrix}3&4&-3&-1\\0&-4&4&-2\\0&8&-10&1\\0&0&2&1\end{vmatrix}=\begin{vmatrix}3&4&-3&-1\\0&-4&4&-2\\0&0&-2&-3\\0&0&2&1\end{vmatrix}=3\begin{vmatrix}-4&4&-2\\0&-2&-3\\0&2&1\end{vmatrix}=3\cdot (-4)\begin{vmatrix}-2&-3\\2&1\end{vmatrix}=-12\cdot 4=-48$
Note that we expand down the first column of the $4\times 4$ matrix, and then down the first column of the $3\times 3$ matrix.
