Solution
= Solution
The third row is twice the first, so the rows are <linearly dependent>. Hence
$$
\boxed{\begin{vmatrix}4&1&-2\\3&0&4\\8&2&-4\end{vmatrix}=0.}
$$
= Solution
The third row is twice the first, so the rows are <linearly dependent>. Hence
$$
\boxed{\begin{vmatrix}4&1&-2\\3&0&4\\8&2&-4\end{vmatrix}=0.}
$$