How does Laplace expansion work? 
\begin{bmatrix} 1 & 2 & 0 & 0 & a\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1 & 0\\ 1 & 0 & 1 & 1 & 1 \end{bmatrix}
Caculate its determinant.

Although it has small numbers, it's still fairly long to calculate the determinant of a large matrix. Peeking at the solution to this exercise, it says

Applying Laplace expansion on the second row and then on the third, we
  obtain a matrix of size $3$ where we can apply Sarrus. The determinant
  is $1 - a$.

I don't really know either of those terms, Laplace and Sarrus. Looking up at Wikipedia, Sarrus is pretty clear to me now, but Laplace is still fairly confusing.
How does Laplace expansion work?
 A: Hint: Expand along the rows that have the most zeros.
Expanding along row 2, we have:
$$\begin{vmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1  & 1 & 1 & 1 \end{vmatrix}$$
Note, what I mean by expanding along row $2$ using Laplace Expansion (see link below), is:
$$0 \begin{vmatrix} 2 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 0  & 1 & 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1  & 1 & 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 & 0 & a\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1  & 0 & 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 & 0 & a\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\\ 1  & 0 & 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 & 0 & 0\\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\\ 1  & 0 & 1 & 1 \end{vmatrix}$$  
Notice all of the zero terms that are leading? Only one item with a $1$ in front survives. Repeating this process two more times yields the result as follows.
Expanding along row 3, we have:
$$\begin{vmatrix} 1 & 0 & a\\ 0 & 1 & 0 \\ 1  & 1 & 1 \end{vmatrix}$$
Expanding along row 2, we have:
$$\begin{vmatrix} 1 & a\\ 1 & 1 \end{vmatrix}$$
Result:
$$\mbox{det} ~ = ~ (1-a)$$
I am basically using Laplace Expansion, because all the leading numbers for the expansion are zero. The site includes an example of where this is not the case.
