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let $A$, $B$ and $C$ be three matrices, such that $A$ and $B$ can be multiplied, $A$ and $C$ can also be multiplied, and we can add $B$ to $C$.

Prove that $$A(B+C)=AB+AC$$

This is my proof (it's probably wrong.)

since we can add $B$ to $C$ this implies that if $B:n\times s$ then $C:n\times s$ and $A:m\times n$

let $$\cases{A=(a_{ij}) \\ B=(b_{jk}) \\ C=(c_{jk})} \implies B+C=(b_{jk}+c_{jk}), \text{ } i=1,..,m\text{ and } j=1,..,n$$ and using the definition of matrix multiplication $$ A(B+C)= \sum_{j=1}^n a_{ij}(b_{jk}+c_{jk})=\sum_{j=1}^n a_{ij}b_{jk}+\sum_{j=1}^n a_{ij}c_{jk}$$ $$=AB+AC$$

I've interpreted $a_{ij}$ and $b_{jk}$ as numbers so that I can apply the distributive rule between them, but i suspect that $a_{ij}$ and $b_{jk}$ are matrices, so I can't do that, is there any method to make this proof rigorous enough?

PNT
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  • You're missing a plus sign between the latter sums but otherwise it looks pretty good, look @Siong's answer – LegNaiB May 17 '21 at 10:51
  • Everything's correct except some notation. You should use $A(B+C)_{ij}$ at LHS instead of $A(B+C)$, since you're comparing the entries. – Lab May 17 '21 at 11:00

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You are comparing the $(i,k)$ entry.

$(i,k)$entry of $A(B+C)$ is $\sum_{j=1}^n a_{ij} (b_{jk}+c_{jk})$.

$(i,k)$entry of $AB+AC$ is $\sum_{j=1}^n a_{ij}b_{jk}+\sum_{j=1}^na_{ij}c_{jk}$.

Note that $a_{ij}$ and $b_{jk}$ are numbers.

Siong Thye Goh
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  • if $a_{ij}$ and $b_{jk}$ are just numbers then $$AB=\sum_{j=1}^n a_{ij}b_{jk}=\sum_{j=1}^n b_{jk}a_{ij}=BA?$$ @SiongThyeGoh – PNT May 17 '21 at 12:20
  • $AB$ is a matrix. It's $(i,k)$ entry is $\sum_{j=1}^n a_{ij}b_{jk}$ is a number. Also be very careful about the definition of matrix multiplication. Try to write down the $(i,k)$ entry of $BA$. – Siong Thye Goh May 17 '21 at 12:23
  • I got it, I thought that the $(i,k)$ entry is the same thing as the matrix, now if we want to prove that $A(B+C)=AB+AC$ we just have to prove that the entries are the same? @SiongThyeGoh – PNT May 17 '21 at 13:23
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    that's the meaning of two matrices are equal, each of the components are equal. – Siong Thye Goh May 17 '21 at 13:29