Question

Let $\mathit{A}\mathbf{=}\mathbf{\left(}{\mathbf{a}}_{\mathbf{i}\mathbf{,}\mathbf{j}}\mathbf{\right)}$be an $\mathit{n}\mathbf{\times }\mathit{m}$matrix, $\mathit{B}\mathbf{=}\mathbf{\left(}{\mathbf{b}}_{\mathbf{i}\mathbf{,}\mathbf{j}}\mathbf{\right)}$be a $\mathit{m}\mathbf{\times }\mathit{k}$ matrix, and $\mathit{C}\mathbf{=}\mathbf{\left(}{\mathbf{c}}_{\mathbf{i}\mathbf{,}\mathbf{j}}\mathbf{\right)}$be a $\mathit{k}\mathbf{\times }\mathit{p}$ matrix. Prove that$\mathit{A}\mathbf{\left(}\mathbf{B}\mathbf{C}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{A}\mathbf{B}\mathbf{\right)}\mathit{C}$ . [Hint: $\mathit{B}\mathit{C}\mathbf{=}{\mathit{d}}_{\mathbf{i}\mathbf{,}\mathbf{j}}$, where ${\mathit{d}}_{\mathbf{i}\mathbf{,}\mathbf{j}}\mathbf{=}\sum _{r=1}^k{b}_{i,r}{c}_{r,j}$and $\mathit{A}\mathit{B}\mathbf{=}{\mathit{r}}_{\mathbf{i}\mathbf{,}\mathbf{r}}$, where${\mathit{e}}_{\mathbf{i}\mathbf{,}\mathbf{r}}\mathbf{=}\mathbf{\sum }_{\mathbf{t}\mathbf{=}\mathbf{1}}^{\mathbf{m}}{\mathbf{a}}_{\mathbf{i}\mathbf{,}\mathbf{t}}{\mathbf{c}}_{\mathbf{t}\mathbf{,}\mathbf{r}}$ . The i-j entry of$\mathit{A}\mathbf{\left(}\mathbf{B}\mathbf{C}\mathbf{\right)}$ is$\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{1}}^{\mathbf{k}}{\mathit{a}}_{\mathbf{i}\mathbf{,}\mathbf{t}}{\mathit{d}}_{\mathbf{t}\mathbf{,}\mathbf{j}}\mathbf{=}\sum _{t=1}^m{a}_{i,t}\left(\sum _{r=1}^k{b}_{i,r}{c}_{r,j}\right)\mathbf{=}\sum _{t=1}^m\sum _{r=1}^k{a}_{i,t}{b}_{i,r}{c}_{r,j}$ . Show that the i-j entry of $\mathbf{\left(}\mathbf{A}\mathbf{B}\mathbf{\right)}\mathit{C}$is this same double sum.]

Short answer

The required identity$A\left(BC\right)=\left(AB\right)C$ has been proved.

Step-by-step solution

Find the matrix sum of  A(BC)

It is given that the$A\mathbf{=}\left(a_{i,j}\right)$be an$n\times m$matrix,$B\mathbf{=}\left(b_{i,j}\right)$be a$m\times k$matrix, and$C\mathbf{=}\left(c_{i,j}\right)$be a$k\times p$matrix.

For the left-side term, then the matrix product of$A\left(BC\right)$is;

$\begin{array}{c}A\left(BC\right)\mathbf{=}A\left(D\right)\\ \mathbf{=}{a}_{i,j}{d}_{tj}\end{array}$

For the right-side term, then the matrix product of role="math" localid="1659422811096" $\left(AB\right)C$ is;

$\begin{array}{c}\left(AB\right)C\mathbf{=}\left(E\right)C\\ \mathbf{=}{e}_{i,r}{c}_{ij}\end{array}$

Note that for each $(i,j)$, $d_{i,j}\mathbf{=}\sum _{r\mathbf{=}1}^k{b}_{i,r}{c}_{r,j}$ and similarly for, $e_{i,r}\mathbf{=}\sum _{t\mathbf{=}1}^m{a}_{i,t}{c}_{t,r}$.

Find the values of ai,jdtjandei,rcijei,rcij.

Now, take left side, the$i\mathbf{-}j$entry of$A\left(BC\right)$,

$\begin{array}{c}\sum _{r\mathbf{=}1}^k{a}_{i,t}{d}_{t,j}\mathbf{=}\sum _{t\mathbf{=}1}^m{a}_{i,t}\left(\sum _{r\mathbf{=}1}^k{b}_{i,r}{c}_{r,j}\right)\\ \mathbf{=}\sum _{t\mathbf{=}1}^m\sum _{r\mathbf{=}1}^k{a}_{i,t}{b}_{i,r}{c}_{r,j}\end{array}$

Now, take left side, $i\mathbf{-}j$theentry of$\left(AB\right)C$,

$\begin{array}{c}\sum _{r\mathbf{=}1}^k{e}_{i,r}{c}_{r,j}\mathbf{=}\left(\sum _{t\mathbf{=}1}^m{a}_{i,t}{b}_{t,r}\right)\sum _{r\mathbf{=}1}^k{c}_{r,j}\\ \mathbf{=}\sum _{t\mathbf{=}1}^m\sum _{r\mathbf{=}1}^k{a}_{i,t}{b}_{i,r}{c}_{r,j}\end{array}$

Since the both sides are equal. Hence, it can be concluded that the property of associative of product matrix is $A\left(BC\right)=\left(AB\right)C$.