Question

a) Let $\mathit{E}$and $\mathit{F}$be$\mathbf{1}\mathbf{\times }\mathit{m}$ row vectors and $\mathit{G}\mathbf{=}\mathbf{\left(}{\mathbf{g}}_{\mathbf{i}\mathbf{j}}\mathbf{\right)}$an $\mathit{m}\mathbf{\times }\mathit{k}$matrix. Prove that

$\mathbf{(}\mathbf{E}\mathbf{+}\mathbf{F}\mathbf{)}\mathit{G}\mathbf{=}\mathit{E}\mathit{G}\mathbf{+}\mathit{F}\mathit{G}$

(b)Let $\mathit{E}\mathbf{=}\mathbf{\left(}{\mathbf{e}}_{\mathbf{i}\mathbf{,}\mathbf{j}}\mathbf{\right)}$ and $\mathit{F}\mathbf{=}\mathbf{\left(}{\mathbf{f}}_{\mathbf{i}\mathbf{,}\mathbf{j}}\mathbf{\right)}$be $\mathit{n}\mathbf{\times }\mathit{m}$ row vectors and$\mathit{G}\mathbf{=}\mathbf{\left(}{\mathbf{g}}_{\mathbf{i}\mathbf{j}}\mathbf{\right)}$ an$\mathit{m}\mathbf{\times }\mathit{k}$matrix. Prove that$\mathbf{(}\mathbf{E}\mathbf{+}\mathbf{F}\mathbf{)}\mathit{G}\mathbf{=}\mathit{E}\mathit{G}\mathbf{+}\mathit{F}\mathit{G}$ .

Short answer

a) The required identity $(E\mathbf{+}F)G\mathbf{=}EG\mathbf{+}FG$has been proved.

b) The required identity $(E\mathbf{+}F)G\mathbf{=}EG\mathbf{+}FG$ has been proved.

Step-by-step solution

Prove the identity (a) (E+F)G=EG+FG  

It is given that the$E$and$F$be$1\mathbf{\times }m$matrices whose with elements indexed$e_{i,j}$and$f_{i,j}$respectively and$G\mathbf{=}\left(g_{ij}\right)$be an$m\mathbf{\times }k$matrix.

For the left side term, then the matrix sum of $(E\mathbf{+}F)G$is,

$\begin{array}{c}(E\mathbf{+}F)G\mathbf{=}(E\mathbf{+}F)\cdot g_{i,j}\\ \mathbf{=}E\cdot g_{i,j}\mathbf{+}F\cdot g_{i,j}\\ \mathbf{=}EG\mathbf{+}FG\end{array}$

For the right-side term, then the matrix sum of $EG\mathbf{+}FG$ is,

$\begin{array}{c}EG\mathbf{+}FG\mathbf{=}E\cdot g_{i,j}\mathbf{+}F\cdot g_{i,j}\\ \mathbf{=}(E\mathbf{+}F)\cdot g_{i,j}\\ \mathbf{=}(E\mathbf{+}F)G\end{array}$

Hence, it is proved that the property of distributive with scalar addition matrix is$(E\mathbf{+}F)G\mathbf{=}EG\mathbf{+}FG$.

Prove the identity  (b) (E+F)G=EG+FG 

It is given that the$E$and$F$be$n\mathbf{\times }m$matrices whose with elements indexed$e_{i,j}$and$f_{i,j}$respectively that is$E\mathbf{=}\left(e_{ij}\right)$, and$F\mathbf{=}\left(f_{ij}\right)$, and$G\mathbf{=}\left(g_{ij}\right)$be a$m\mathbf{\times }k$matrix.

For left side term, then the matrix sum of$(E\mathbf{+}F)G$ is,

$\begin{array}{c}(E\mathbf{+}F)G\mathbf{=}(e_{ij}\mathbf{+}{f}_{ij})\cdot g_{i,j}\\ \mathbf{=}{e}_{ij}\cdot g_{i,j}\mathbf{+}{f}_{ij}\cdot g_{i,j}\\ \mathbf{=}EG\mathbf{+}FG\end{array}$

For right-side term, then the matrix sum of$EG\mathbf{+}FG$ is,

$\begin{array}{c}EG\mathbf{+}FG\mathbf{=}{e}_{ij}\cdot g_{i,j}\mathbf{+}{f}_{ij}\cdot g_{i,j}\\ \mathbf{=}(e_{ij}\mathbf{+}{f}_{ij})\cdot g_{i,j}\\ \mathbf{=}(E\mathbf{+}F)G\end{array}$

Hence, it is proved that the property of distributive with scalar addition matrix is $(E\mathbf{+}F)G\mathbf{=}EG\mathbf{+}FG$ .