Let - $K^{m\times n}\ni A=\left(a_{i j}\right)$ - $K^{n\times r}\ni B=\left(b_{j k}\right)$ - $K^{r\times s}\ni C=\left(c_{k l}\right)$ be [[matrix|matrices]] such that - $A, B$ can be [[matrix multiplication|multiplied]] - $B, C$ can be [[matrix multiplication|multiplied]] Then - $A, B C$ can be [[matrix multiplication|multiplied]] - $A B, C$ can be [[matrix multiplication|multiplied]] and $ (A B) C=A(B C) $ #### Proof $AB$ is $m\times r$, and the $(i,k)$ element is a dot product of row $i$ of $A$ and column $k$ of $B$ $AB=\left(\begin{array}{ccc} \mathbf a_{\text{r } 1}\cdot \mathbf b_{\text{c }1}&\ldots&\mathbf a_{\text{r } 1}\cdot \mathbf b_{\text{c }k}&\ldots&\mathbf a_{\text{r } 1}\cdot \mathbf b_{\text{c }r}\\ \vdots&\ddots&\vdots&\ddots&\vdots\\ \mathbf a_{\text{r } i}\cdot \mathbf b_{\text{c }1}&\ldots &\mathbf a_{\text{r } i}\cdot \mathbf b_{\text{c }k}&\ldots&\mathbf a_{\text{r } i}\cdot \mathbf b_{\text{c }r}\\ \vdots&\ddots&\vdots&\ddots&\vdots\\ \mathbf a_{\text{r } m}\cdot \mathbf b_{\text{c }1}&\ldots&\mathbf a_{\text{r } m}\cdot \mathbf b_{\text{c }k}&\ldots&\mathbf a_{\text{r } m}\cdot \mathbf b_{\text{c }r} \end{array}\right)$ Multiplying this $AB$ matrix by $C$ (on the right), results in $(AB)C,$ an $m\times s$ matrix. $(AB)C=\left(\begin{array}{ccc} (\mathbf{ab})_{\text{r }1}\cdot\mathbf c_{\text{c }1}&\ldots& (\mathbf{ab})_{\text{r }1}\cdot\mathbf c_{\text{c }l}&\ldots&(\mathbf{ab})_{\text{r }1}\cdot\mathbf c_{\text{c }s}\\ \vdots&\ddots&\vdots&\ddots&\vdots\\ (\mathbf{ab})_{\text{r }i}\cdot\mathbf c_{\text{c }1}&\ldots& (\mathbf{ab})_{\text{r }i}\cdot\mathbf c_{\text{c }l}&\ldots&(\mathbf{ab})_{\text{r }i}\cdot\mathbf c_{\text{c }s}\\ \vdots&\ddots&\vdots&\ddots&\vdots\\ (\mathbf{ab})_{\text{r }m}\cdot\mathbf c_{\text{c }1}&\ldots& (\mathbf{ab})_{\text{r }m}\cdot\mathbf c_{\text{c }l}&\ldots&(\mathbf{ab})_{\text{r }m}\cdot\mathbf c_{\text{c }s} \end{array}\right)$ So the $i,l$ element in $(AB)C$ is a double sum, $(\mathbf{ab})_{\text{r }i}\cdot\mathbf c_{\text{c }l}=\sum_{k=1}^{r}\left[\sum_{j=1}^{n} a_{i j} b_{j k}\right] c_{k l}=\sum_{k=1}^{r}\left[\sum_{j=1}^{n} a_{i j} b_{j k} c_{k l}\right]$ Taking similar steps for $A(BC)$, $\mathbf{a}_{\text{r }i}\cdot (\mathbf{bc})_{\text{c }l}=\sum_{j=1}^{n}\left[a_{i j}\sum_{k=1}^{r} b_{j k}c_{k l}\right] =\sum_{j=1}^{n}\left[\sum_{k=1}^{r} a_{i j} b_{j k} c_{k l}\right]$ And both double sums are equivalent to $ \sum_{\tiny\begin{array}{c}1 \leqq j \leqq n\\1 \leqq k \leqq r\end{array}}a_{i j} b_{j k} c_{k l} $ Since every element of $A(BC)$ is equal to $(AB)C$, the matrices themselves are equal: $A(BC)=(AB)C$