Question

Calcula la matriz X



2A − 3(A+B) = 3X − 2(A−B)

Answer 1

Para resolver este problema recordemos lo siguiente

              $\left[\begin{array}{cccc}\sf{a}&\sf{b}&\sf{c}\\\sf{d}&\sf{e}&\sf{f}\\\sf{g}&\sf{h}&\sf{i}\\\end{array}\right]+\left[\begin{array}{cccc}\sf{m}&\sf{n}&\sf{o}\\\sf{p}&\sf{q}&\sf{r}\\\sf{s}&\sf{t}&\sf{u}\\\end{array}\right]=\left[\begin{array}{cccc}\sf{a+m}&\sf{b+n}&\sf{c+o}\\\sf{d+p}&\sf{e+q}&\sf{f+r}\\\sf{g+s}&\sf{h+t}&\sf{i+u}\\\end{array}\right]$

              $\left[\begin{array}{cccc}\sf{a}&\sf{b}&\sf{c}\\\sf{d}&\sf{e}&\sf{f}\\\sf{g}&\sf{h}&\sf{i}\\\end{array}\right]-\left[\begin{array}{cccc}\sf{m}&\sf{n}&\sf{o}\\\sf{p}&\sf{q}&\sf{r}\\\sf{s}&\sf{t}&\sf{u}\\\end{array}\right]=\left[\begin{array}{cccc}\sf{a-m}&\sf{b-n}&\sf{c-o}\\\sf{d-p}&\sf{e-q}&\sf{f-r}\\\sf{g-s}&\sf{h-t}&\sf{i-u}\\\end{array}\right]$

                    $\sf{m}\times\left[\begin{array}{cccc}\sf{a}&\sf{b}&\sf{c}\\\sf{d}&\sf{e}&\sf{f}\\\sf{g}&\sf{h}&\sf{i}\\\end{array}\right]=\left[\begin{array}{cccc}\sf{m\times a}&\sf{m\times b}&\sf{m\times c}\\\sf{m\times d}&\sf{m\times e}&\sf{m\times f}\\\sf{m\times g}&\sf{m\times h}&\sf{m\times i}\\\end{array}\right]$

Entonces en el problema nos piden

$\begin{array}{c}\boldsymbol{\sf{2A-(3A+B)=3X-2(A-B)}}\\\\\sf{2\left[\begin{array}{cccc}\sf{8}&\sf{-7}\\\sf{-5}&\sf{9}\\\end{array}\right]-\left(3\left[\begin{array}{cccc}\sf{8}&\sf{-7}\\\sf{-5}&\sf{9}\\\end{array}\right]+\left[\begin{array}{cccc}\sf{3}&\sf{2}\\\sf{1}&\sf{10}\\\end{array}\right]\right)=3X-2\left(\left[\begin{array}{cccc}\sf{8}&\sf{-7}\\\sf{-5}&\sf{9}\\\end{array}\right]-\left[\begin{array}{cccc}\sf{3}&\sf{2}\\\sf{1}&\sf{10}\\\end{array}\right]\right)}\\\end{array}$

   $\sf{\left[\begin{array}{cccc}\sf{16}&\sf{-14}\\\sf{-10}&\sf{18}\\\end{array}\right]-\left(\left[\begin{array}{cccc}\sf{24}&\sf{-21}\\\sf{-15}&\sf{27}\\\end{array}\right]+\left[\begin{array}{cccc}\sf{3}&\sf{2}\\\sf{1}&\sf{10}\\\end{array}\right]\right)=3X-2\left(\left[\begin{array}{cccc}\sf{5}&\sf{-9}\\\sf{-6}&\sf{-1}\\\end{array}\right]\right)}$

               $\sf{\left[\begin{array}{cccc}\sf{16}&\sf{-14}\\\sf{-10}&\sf{18}\\\end{array}\right]-\left[\begin{array}{cccc}\sf{27}&\sf{-19}\\\sf{-14}&\sf{37}\\\end{array}\right]=3X-\left(\left[\begin{array}{cccc}\sf{10}&\sf{-18}\\\sf{-12}&\sf{-2}\\\end{array}\right]\right)}$

                             $\sf{\left[\begin{array}{cccc}\sf{-11}&\sf{5}\\\sf{4}&\sf{-19}\\\end{array}\right]=3X-\left[\begin{array}{cccc}\sf{10}&\sf{-18}\\\sf{-12}&\sf{-2}\\\end{array}\right]}$

                            $\sf{3X=\left[\begin{array}{cccc}\sf{-11}&\sf{5}\\\sf{4}&\sf{-19}\\\end{array}\right]+\left[\begin{array}{cccc}\sf{10}&\sf{-18}\\\sf{-12}&\sf{-2}\\\end{array}\right]}$

                                            $\sf{3X=\left[\begin{array}{cccc}\sf{-1}&\sf{-13}\\\sf{-8}&\sf{-21}\\\end{array}\right]}$

                                          $\sf{X=\left[\begin{array}{cccc}\sf{-1/3}&\sf{-13/3}\\\sf{-8/3}&\sf{-21/3}\\\end{array}\right]}$

                                   $\boxed{\boxed{\boldsymbol{\red{\vphantom{\Bigg|}\hphantom{a}\sf{X=\left[\begin{array}{cccc}\sf{-1/3}&\sf{-13/3}\\\sf{-8/3}&\sf{-7}\\\end{array}\right]\hphantom{a}}}}}}$

                                            $\boxed{\sf{{R}}\quad\raisebox{10pt}{ \sf{\red{O}} }\!\!\!\!\raisebox{-10pt}{ \sf{\red{O}} }\quad\raisebox{15pt}{ \sf{{G}} }\!\!\!\!\raisebox{-15pt}{ \sf{{G}} }\quad\raisebox{15pt}{ \sf{\red{H}} }\!\!\!\!\raisebox{-15pt}{ \sf{\red{H}} }\quad\raisebox{10pt}{ \sf{{E}} }\!\!\!\!\raisebox{-10pt}{ \sf{{E}} }\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}$