Steps on proving $(\det A)(\det B)=\det(AB)$

Question

I am going through the steps on proving $det(AB)=det(A)det(B)$. Let $A=(u_1 \dots u_n)$,$B=(v_1 \dots v_n)$ let $e_{i_{k}}$ denote the $n \times 1$ column vector with zeros in every row but a 1 in the kth slot.If we let $A=(a_{j,k})_{1 \leq j,k \leq n}$ $B=(b_{j,k})_{1 \leq j,k \leq n}$ $A$ and $B$ are $n \times n$ matrices

Prove $Av_k=$$\sum_{i_{k}=1}^{n} b_{i_k,k}Ae_{i_k}$

I have a feeling that I need to use matrix multiplication to prove this.

Or could I just see that $v_k=\sum_{i_{k}=1}^{n} b_{i_k,k}e_{i_k}$

and say $Av_k=$$\sum_{i_{k}=1}^{n} b_{i_k,k}Ae_{i_k}$

If not how would I show this using matrix multiplication?

Answer 1

Yes, what you wrote is fine. These are the reasonings.

\begin{align} Av_k &= A\left( \sum_{i_k=1}^n b_{i_k, k}e_{i_k} \right) \\ &=\sum_{i_k=1}^n Ab_{i_k, k}e_{i_k} \text{, this is due to distributive law}\\ &=\sum_{i_k=1}^n b_{i,k, k} Ae_{i_k}, \text{ since } b_{i,k} \text{ is a scalar.} \end{align}