Computing the $4 \times 4$ determinant of a matrix 
Compute the determinant of
$$\begin{bmatrix}1&-2&5&2\\0&0&3&0\\2&-4&-3&5\\2&0&3&5\end{bmatrix}$$
by first expanding along the first row (at every stage) and then by expanding along whatever row or column requires the fewest computations

I'm trying to put this matrix in Lower Triangular Form (LTF) so that I can find the determinant by just taking the product of the diagonal. I'm stuck though, I'm not sure where I went wrong.
I did:
$$R_3 = 2R_1 + (-1)R_3$$
$$R_4 = 2R_1 + (-1)R_4$$
$$R_4 = 7R_1 + (-1)5R_4$$
to get:
$$\begin{bmatrix}1&-2&5&2\\0&0&3&0\\0&0&13&-1\\7&6&0&19\end{bmatrix}$$
 A: What they want is an "expansion by minors" which is given by the formula 
$$
|A|=\sum_i^j(-1)^{i+j}a_{ij}M_{ij}
$$
Which is more intimidating looking than the process actually is. You pick a row or column and go across the row or column and for each element $a_{ij}$, "delete" the i'th row and j'th column in your mind, finding the determinant of the submatrix $M_{ij}$. 
Keeping track of the sign of each term of your sum determined by the parity of $i+j$, you sum each mini determinant to get the big determinant. 
In your case you will have to do expand again within your expansion to get down to a 2 by 2 block, where you can use the formula $ad-bc$.
If you want to do it your way, I suppose you could row reduce to a upper triangular matrix, keeping track of your operations; row swapping changes the sign of the determinant, multiplication of a row by a constant multiplies your matrix by a constant (the determinant is linear in columns and rows). That being said, you should learn cofactor expansion, it's important. 
A: \begin{eqnarray}
\left|\begin{matrix}1&-2&5&2\\0&0&3&0\\2&-4&-3&5\\2&0&3&5\end{matrix}\right|&=& -3\left|\begin{matrix}1&-2&2\\2&-4&5\\2&0&5\end{matrix}\right|\\
&=& -3 \left|\begin{matrix}1&0&2\\2&0&5\\2&4&5\end{matrix}\right|\\
&=& -3\times(-4) \left|\begin{matrix}1&2\\2&5\end{matrix}\right|\\
&=& 12\times (5-4)\\
&=&12
\end{eqnarray}
where in the first step i expand the determinant from the second line (because they have 3 zero in this line) and i remove the second line and the third column.
for the second step i "create" more zero by adding to the second column two times the first column and i expand with this column to get the last $2\times 2 $ determinant which calculation is very easy 
A: "expanding the determinant" means doing
$$
||M|| = 1\cdot \left|\left| \matrix{0&3&0\\-4&-3&5\\0&3&5}\right|\right|-(-2)\cdot
\left|\left| \matrix{0&3&0\\2&-3&5\\2&3&5}\right|\right|+
5\cdot
\left|\left| \matrix{0&0&0\\2&-4&5\\2&0&5}\right|\right|-2\cdot
\left|\left| \matrix{0&0&3\\2&-4&-3\\2&0&3}\right|\right|
$$
