Determinant of $4\times4$ Matrix I tried to solve for a $4 \times 4$ matrix, but I'm unsure if I did this properly, can anyone tell me if I did this correct? Or if there were any mistakes where at? Also, I know this is an inefficient method for finding the determinant, however I want to get practice with solving like so:
$$A=
\begin{bmatrix}
2 & 4 & 0 & 1 \\
0 & 8 & 0 & 2 \\
0 & 3 & 0 & 5 \\
1 & 2 & 1 & 1
\end{bmatrix}
$$

$$\begin{align*}
\det(A)&=2
\begin{vmatrix}
8 & 0 & 2 \\
3 & 0 & 5 \\
2 & 1 & 1
\end{vmatrix}
-4
\begin{vmatrix}
0 & 0 & 2 \\
0 & 0 & 5 \\
1 & 1 & 1
\end{vmatrix}
+0
-1
\begin{vmatrix}
0 & 8 & 0 \\
0 & 3 & 0 \\
1 & 2 & 1
\end{vmatrix}\\[0.3in]
&=2\left(8
\begin{vmatrix}
0 & 5 \\
1 & 1
\end{vmatrix}-0+2
\begin{vmatrix}
3 & 0\\
2 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}-4\left(0-0+2
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}\right)\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1\left(0-8
\begin{vmatrix}
0 & 0\\
1 & 1
\end{vmatrix}+0\right)\\[0.3in]
&=2(8(0-5)-0+2(3-0))\\[0.1in]
&\quad{}-4(0-0+2(0))\\[0.1in]
&\quad{}+0\\[0.1in]
&\quad{}-1(0-8(0)+0)\\[0.3in]
&= 2(8(-5)-0+2(3))\\[0.3in]
&=2(-45+6)\\[0.3in]
&=2(-39)\\[0.3in]
&=-78
\end{align*}$$

Sorry for the long post, I tried to make the readability easy for everyone.

I know you said that you know that this is not the most efficient way of finding the determinant, but may I point out a quick way? In fact, you use the same technique, but apply it to the third column.
$$\det(A) = (-1)^{3+4} 
\left| \begin{array}{lll} 
2 & 4 & 1 \\
0 & 8 & 2 \\
0 & 3 & 5
\end{array}\right| = -2 \left| \begin{array}{lll} 
8 & 2 \\
3 & 5
\end{array}\right| = -68
 $$
