Breaking a determinant into eight pieces From Strang, Introduction to Linear Algebra, 4th ed. p. 250. This is an example whose answer is given in the text but I'm confused. I've paraphrased and corrected what I think is a typo in the first matrix (row 3 + row 3 changed to row 3 + row 2).
How is the determinant of M related to the determinant of A? Rows of M are combinations of the rows of A.
$$
M=
\begin{bmatrix}  \text{row 1 + row 3} \\ \text{row 2 + row 1} \\ \text{row 3 + row 2} \\  \end{bmatrix} 
$$
M can be split into eight matrices:
$$
\det M=
\begin{vmatrix}  \text{row 1 + row 3} \\ \text{row 2 + row 1} \\ \text{row 3 + row 2} \\  \end{vmatrix} 
=  \begin{vmatrix} \text{row 1} \\ \text{row 2} \\ \text{row 3} \end{vmatrix}
+ \begin{vmatrix} \text{row 3} \\ \text{row 2} \\ \text{row 3} \end{vmatrix}
+ \begin{vmatrix} \text{row 1} \\ \text{row 1} \\ \text{row 3} \end{vmatrix}
+ \cdots 
+ \begin{vmatrix} \text{row 3} \\ \text{row 1} \\ \text{row 2} \end{vmatrix} 
$$ 
All but the first and last have repeated rows and zero determinant.
My question: By the rule that $$\begin{vmatrix} a + a^\prime  & b + b^\prime \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a^\prime  & b^\prime \\ c & d \end{vmatrix}$$ I'd have thought that the first three matrices (before the ellipsis) would suffice to produce the determinant of M. Why do we need the other bunch?

Let's apply the rule you quoted once:
$$
\det M=\left|
\begin{matrix}
\text{row 1}+\text{row 3}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|
=
\left|
\begin{matrix}
\text{row 1}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|+
\left|
\begin{matrix}
\text{row 3}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|
$$
Now you apply the rule to the second row, and it gives $2\cdot 2=4$ terms. Lastly apply it to the third row to get all $2\cdot 2\cdot 2=8$ terms.
Each time you apply the rule to a row, the number of terms doubles.
