I have looked over this problem five times and I am not getting the right answer so I was wondering if anybody can perhaps spot the error.
$$A=\begin{bmatrix}4 & 0 & 0 & 5 \\ 1 & 7 & 2 & -5 \\ 3 & 0 & 0 & 0 \\ 8 & 3 & 1 & 7 \end{bmatrix}$$
$$\det(A) = 5?$$
I am getting it's supposed to be $15$ buy I'm not sure where I erred.
So I choose to expand across row 3 so I get
$$B=\begin{bmatrix} & 0 & 0 & 5 \\ & 7 & 2 & -5 \\ & 3 & 1 & 7 \end{bmatrix}$$
$$\det(A) = 3 \det B$$
Then finally if you expand across the first row, only the last column should matter because all the other terms go to 0, right? In which case you'd get
$$C=\begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix}$$
$\det(A) = 5 \det C = 5(7-6) = 5$?
You wrote that the $\det(A) = 3 \det(B)$, which you would be correct on.
Though, the $\det(B) = 5\det(C)$ where
$C = \begin{bmatrix} 7&2\\ 3&1 \end{bmatrix} $
In this case $\det(C) = 7(1)-2(3) = 1$.
Therefore, $\det(A) = 3(\det(B)) = 3(5\det(C)) = 3(5)(1) = 15$.