The product of $100\times 2$ and $2\times 100$ matrices must have zero determinant

Question

Let A be a 100 × 2 matrix and let B be a 2 × 100 matrix. Then C = AB is a 100 × 100 matrix. Explain why $\det(C) = 0$.

My initial thoughts on this question is that if $\det(C)=0$, then it cannot be expressed as the product of elementary matrices, which would be the case given that both A and B are not square, and hence they themselves cannot be expressed this way.

Is my line of thinking with this correct, or am I missing something glaringly obvious?

Answer 1

Hint:

1. $rank(A_{m\times n})\le\min\{m,n\}=2$ (in this case)
2. $rank(A_{m\times n}B_{n\times m})\le\min\{rank(A),rank(B)\}\le2$ (in this case)

and 3. if $rank(A_{n\times n})< n$, then $\det(A)=0$.

Edit: Since OP is yet to be introduced to the term "rank", adding from comments:

Let $\bar A_{100\times100}=\left[\begin{matrix} A_{100\times2} && 0_{100\times98}\end{matrix}\right]$, $\bar B_{100\times100}=\left[\begin{matrix} B_{2\times100} \\ 0_{98\times100}\end{matrix}\right]$

If $A=(a_{i,j})_{100\times2}$, $B=(b_{i,j})_{2\times100}$,

$\bar A$ will look like

$$\left[\begin{matrix} a_{1,1} && a_{1,2} && 0 && \cdots && 0 \\ a_{2,1} && a_{2,2} && 0 && \cdots && 0\\ \vdots && \vdots && \vdots && \vdots && \vdots &&\\ a_{100,1} && a_{100,2} && 0 && \cdots && 0\end{matrix}\right]$$

and $\bar B$ will look like

$$\left[\begin{matrix} b_{1,1} && b_{1,2} && \cdots && b_{1,100} \\ b_{2,1} && b_{2,2} && \cdots && b_{2,100} \\ 0 && 0 && \cdots && 0\\ \vdots && \vdots && \vdots && \vdots &&\\ 0 && 0 && \cdots && 0\end{matrix}\right]$$

Note that $\bar A\bar B=AB$, imagine number of zero rows in $AB$, and effect it will have on determinant.