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.
 A: As pointed out by Amzoti in the comment: your calculations are all correct, save for the glitch at the end, when you erroneously computed $8(-5) = -45$, at the very end, which results in your final calculation being off by $2(-5) = -10$, giving you $-78$ instead of $-68$.  
Correct that, and you're good to go!
But the "hard part" was all perfectly correct. (I double checked the rest of your work, too, and you did just fine.)  so if your point was to "practice" finding the determinant of a matrix by expanding along the first row, to get the process right, you did the key parts well. 
ADDED: If you've learned how elementary row operations alter the determinant of the matrix on which you're operating, doing so can greatly simplify the computation of the determinant of a matrix! See, e.g. Java Man's comment: if you had performed the following elementary row operations $-2R_4 + R_1 \to R_1$, you could have expanded along the first column, greatly simplifying the process. 
If you haven't yet learned how ERO's change the determinant, once you learn that, your work will be greatly reduced in the future! (For example, if you are able to use ERO's on some matrix to reduce it to an equivalent system/matrix containing a row, or a column, of all zeros, the determinant of the entire (original and reduced) matrix will be $0$.
The more practice you get, and the more "shortcuts" you learn, the less tedious computing the determinant of a matrix will become.
A: 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.
$$\det(A) = (-1)^{3+4} 
\left| \begin{array}{lll} 
2 & 4 & 1 \\
0 & 8 & 2 \\
0 & 3 & 5
\end{array}\right| = -2 \left| \begin{array}{lll} 
8 & 2 \\
3 & 5
\end{array}\right| = -68
 $$
