Error computing determinant of a $4\times 4$ matrix I am trying to compute the determinant of 
$$C = \begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ -2 & -2 & -1 & 2 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
I first did the row operation $R_3 \leftarrow R_1-R_3$ so it doesn't change the determinant. So you want the determinant of:
$$ \begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ 0 & 5 & 3 & -1 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
and this simplifies to the $-2$ times the determinant of:
$$C = \begin{bmatrix} 3 & 1 & -3 \\  5 & 3 & -1 \\  2 & 3 & 1  \end{bmatrix}$$
And this $3\times 3$ matrix has determinant of $-16$ so the determinant of $C$ is $(-2)\cdot (-16) = 32$. However, the book says that the answer is $-32$ and not $32$. 
Any hints where the negative comes from?

Recall that  the determinant does not change if you add to a row a linear combination of the others.
You should replace $R_3$ with $R_3-R_1$. Then you obtain
$$\begin{bmatrix}-2 & 3 & 2 & 1 \\ 0 & 3 & 1 & -3 \\ 0 & -5 & -3 & 1 \\ 0 & 2 & 3 & 1 \end{bmatrix}$$
and the determinant of 
$$\begin{bmatrix} 3 & 1 & -3 \\  -5 & -3 & 1 \\  2 & 3 & 1  \end{bmatrix}$$
is just the opposite of $-16$ (your previous computation).
