# Thread: [SOLVED] Multiplication of matrices

1. ## [SOLVED] Multiplication of matrices

The problem has certainly already be solved but I didn't find it in a short research.
Anyway I'm just asking for a check-result.
Prove that if $\displaystyle A$ and $\displaystyle B$ are $\displaystyle n\times m$ matrices and $\displaystyle C$ is a $\displaystyle m\times q$ one, then $\displaystyle (A+B)C=AC+BC$.
My attempt : By definition, $\displaystyle (A+B)_{ij}=A_{ij}+B_{ij}$ so $\displaystyle (A+B)_{ij}C_{ij}=\sum_{k=1}^n [A_{ik}+B_{ik}]C_{kj}$ (call this 1)$\displaystyle =\sum_{k=1}^n A_{ik}C_{kj}+B_{ik}C_{kj}$ (call this 2)$\displaystyle =AC+BC$.
I'm not sure if I can pass from 1 to 2 as I did or I need more calculus/or explanation.

2. Originally Posted by arbolis
The problem has certainly already be solved but I didn't find it in a short research.
Anyway I'm just asking for a check-result.
Prove that if $\displaystyle A$ and $\displaystyle B$ are $\displaystyle n\times m$ matrices and $\displaystyle C$ is a $\displaystyle m\times q$ one, then $\displaystyle (A+B)C=AC+BC$.
My attempt : By definition, $\displaystyle (A+B)_{ij}=A_{ij}+B_{ij}$ so $\displaystyle (A+B)_{ij}C_{ij}=\sum_{k=1}^n [A_{ik}+B_{ik}]C_{kj}$ (call this 1)$\displaystyle =\sum_{k=1}^n A_{ik}C_{kj}+B_{ik}C_{kj}$ (call this 2)$\displaystyle =AC+BC$.
I'm not sure if I can pass from 1 to 2 as I did or I need more calculus/or explanation.
Are $\displaystyle A_{ik},~B_{ik},~C_{kj}\in\mathbb{R}$?

If so, I'm pretty sure it is safe to assume that $\displaystyle \sum_{k=1}^n [A_{ik}+B_{ik}]C_{kj}=\sum_{k=1}^n A_{ik}C_{kj}+B_{ik}C_{kj}=AC+BC$, since we can apply the distributive property to real numbers.

--Chris

3. Are ?
I think so, or $\displaystyle \mathbb{C}$. So that means the demonstration is right.

4. Hi guys,

There is a little problem with that. Just to be picky, but it looks important to me >.<

Let $\displaystyle A=\left[A_{ij}\right]_{\substack{1\le i \le n \\ 1 \le j \le {\color{red}m}}}$
etc...

It is clear (well in fact you gotta know the conventions) that the second coordinate of A_{ij} goes from 1 and m and not 1 and n.

So $\displaystyle (A+B)_{ij}C_{ij}=\sum_{k=1}^n [A_{ik}+B_{ik}]C_{kj}$ is rather $\displaystyle (A+B)_{ij}C_{ij}=\sum_{k=1}^{{\color{red}m}} [A_{ik}+B_{ik}]C_{kj}$

The first coordinate = # of the row.
The second coordinate = # of the column.

A matrix (n,m) is meant to have n rows and m columns.

5. Thank you Moo for pointing that out. If I have something similar in a exam I'll take care of not doing this error.