Math Challenges

Solution:
\( \displaylines{\text{Adding row 2 to rows 3 and 4 give us the same determinant but the matrix is as follows:}\\ \begin{pmatrix}2 & 0 & 0 & 8\\ 1 & -7 & -5 & 0\\ 4 & 1 & 1 & 0\\ 1 & 0 & 0 & 4\end{pmatrix}\\ \text{Expanding across the first row gives a determinant value of}\\ 2\det\begin{pmatrix}-7 & -5 & 0\\ 1 & 1 & 0\\ 0 & 0 & 4\end{pmatrix}-8\det\begin{pmatrix}1 & -7 & -5\\ 4 & 1 & 1\\ 1 & 0 & 0\end{pmatrix}\\ \text{Expanding these across their third row gives zero.}} \)