Row Reduction with Cofactor Expansion My calculator says the determinant of $$\begin{pmatrix}3 &0&6&-3\\0&2&3&0\\-4&-7&2&0\\2&0&1&10\end{pmatrix}$$ is $396$. However, the website I got the matrix from says a number over 1,000, and my own answer is 12. I'm wondering how to correctly arrive at the answer 396. Can a good linear algebraist help?
 A: $$\begin{pmatrix}3 &0&6&\!\!\!\!-3\\0&2&3&0\\\!\!\!\!-4&\!\!\!\!-7&2&0\\2&0&1&10\end{pmatrix}\stackrel{\text{R_4}+\frac{10}{3}\,\text{R_1}}\longrightarrow\begin{pmatrix}3 &0&6&\!\!\!-3\\0&2&3&0\\\!\!-4&\!\!-7&2&0\\12&0&21&0\end{pmatrix}$$
Develop the determinant wrt the fourth column, and I get:
$$\det A=3\,\,\left|\begin{pmatrix}0&2&3\\\!\!-4&\!\!-7&2\\12&0&21\end{pmatrix}\right|=3\left(48+168+252\right)=1,404$$
A: Let us cofactor expand along the last column. 
First let us calculate the determinant of the sub-matrix formed by removing the first row and last column $$M_{14} = \begin{vmatrix}0 & 2& 3 \\ -4 & -7 & 2 \\ 2 & 0 & 1\end{vmatrix}$$
We expand this sub-matrix along the last row
$$\det M_{14} = 2\begin{vmatrix}2 & 3 \\ -7 & 2\end{vmatrix} +\begin{vmatrix}0 & 2 \\ -4 & -7\end{vmatrix}=2(4 + 21) + 8 = 58$$
Now we calculate the determinant of the sub-matrix formed by removing the last row and last column
$$M_{44} = \begin{vmatrix}3 & 0 & 6 \\ 0 & 2& 3 \\ -4 & -7 & 2\end{vmatrix}$$
We expand along the first row to get
$$\det M_{44} = 3\begin{vmatrix}2 & 3 \\ -7 & 2\end{vmatrix} + 6\begin{vmatrix}0 & 2 \\ -4 & -7\end{vmatrix} = 3(4 + 21) + 6(8) = 123$$
Now the full cofactor expansion is given by 
$$3(58) + 10(123) = 1404$$
