Question

Resovla o sistema matricial :$\left \{ {{x+y=-2A+B} \atop {-x+2y=B-A}} \right.$

$A= \left[\begin{array}{ccc}1&1&2\\2&2&1\\3&1&3\end{array}\right] ; b= \left[\begin{array}{ccc}0&0&1\\2&1&0\\0&0&2\end{array}\right]$  

Serio nao to conseguindo fazer esse.. ajudem...!

A resolução Final é x= $\left[\begin{array}{ccc}-1&-1& \frac{-5}{3} \\ \frac{-4}{3} & \frac{-5}{3} &-1\\-3&-1& \frac{-7}{3} \end{array}\right]$
e y=$\left[\begin{array}{ccc}-1&-1& \frac{-4}{3} \\ \frac{-2}{3} & \frac{-4}{3} &-1\\-3&-1& \frac{-5}{3} \end{array}\right]$.

Answer 1

 Olá Ivan,
bom dia!

Da equação I,

$-2A+B=\\\\-2\cdot\begin{bmatrix}1&1&2\\2&2&1\\3&1&3\end{bmatrix}+\begin{bmatrix}0&0&1\\2&1&0\\0&0&2\end{bmatrix}=\\\\\\\begin{bmatrix}-2&-2&-4\\-4&-4&-2\\-6&-2&-6\end{bmatrix}+\begin{bmatrix}0&0&1\\2&1&0\\0&0&2\end{bmatrix}=\\\\\\\begin{bmatrix}-2&-2&-3\\-2&-3&-2\\-6&-2&-4\end{bmatrix}$


 Da equação II,

$B-A=\\\\\begin{bmatrix}0&0&1\\2&1&0\\0&0&2\end{bmatrix}-\begin{bmatrix}1&1&2\\2&2&1\\3&1&3\end{bmatrix}=\\\\\\\begin{bmatrix}-1&-1&-1\\0&-1&-1\\-3&-1&-1\end{bmatrix}$


 Agora é só resolver o sistema (soma),

$\begin{cases}x+y=\begin{bmatrix}-2&-2&-3\\-2&-3&-2\\-6&-2&-4\end{bmatrix}\\\\-x+2y=\begin{bmatrix}-1&-1&-1\\0&-1&-1\\-3&-1&-1\end{bmatrix}\end{cases}\\------------------\\3y=\begin{bmatrix}-3&-3&-4\\-2&-4&-3\\-9&-3&-5\end{bmatrix}\\\\\boxed{y=\begin{bmatrix}-1&-1&\frac{-4}{3}\\\\\frac{-2}{3}&\frac{-4}{3}&-1\\\\-3&-1&\frac{-5}{3}\end{bmatrix}}$

 Ivan, agora acho que consegue encontrar o valor de "x". Caso contrário, pergunte, ok?!