Question

Alguém consegue me ajudar a solucionar esse problema(MATRIZ)??? 

Answer 1

Olá, Mmartinsjr,

$\frac34A+\frac75B+\frac43C+2D=E\Rightarrow\\\\ \frac34\begin{bmatrix} 4 &12 \\ 0 &8 \end{bmatrix}+\frac75\begin{bmatrix} x & 5\\ z & y \end{bmatrix}+\frac43\begin{bmatrix} 6z &15 \\ \frac34x & 3x \end{bmatrix}+2\begin{bmatrix} y &0 \\ 4y &3z \end{bmatrix}=\begin{bmatrix} 6 &36 \\ 1 & 8 \end{bmatrix}\Rightarrow$

$\begin{bmatrix} 3 & 9\\ 0 & 6 \end{bmatrix}+\begin{bmatrix} \frac75x &7 \\\\ \frac75z & \frac75y \end{bmatrix}+\begin{bmatrix} 8z & 20\\ x & 4x \end{bmatrix}+\begin{bmatrix} 2y & 0\\ 8y & 6z \end{bmatrix}=\begin{bmatrix} 6 &36 \\ 1 & 8 \end{bmatrix}\Rightarrow$

$\begin{bmatrix} 3+\frac75x+8z+2y & 9+7+20+0\\ 0+\frac75z+x+8y &6+\frac75y+4x+6z \end{bmatrix}=\begin{bmatrix} 6 &36 \\ 1 & 8 \end{bmatrix}\Rightarrow$

$\left\{\begin{matrix} 3+\frac75x+8z+2y=6 \\36=36\\ \frac75z+x+8y=1 \\6+\frac75y+4x+6z =8 \end{matrix}\right\,\,\,\,\,\,\Rightarrow$

$\left\{\begin{matrix} \frac75x+2y+8z=3 \\\\ x+8y+\frac75z=1 \\\\4x+\frac75y+6z =2 \end{matrix}\right\,\,\,\,\,\,\times5\Rightarrow$

$\left\{\begin{matrix} 7x+10y+40z=15 \\\\ 5x+40y+7z=5 \\\\20x+7y+30z =10 \end{matrix}\right$

$\left\{\begin{matrix} 7x+10y+40z=15 \\\\ -20x-160y-28z=-20 \\\\20x+7y+30z =10 \end{matrix}\right\\\\\\ \left\{\begin{matrix} 7x+10y+40z=15 \\\\ -20x-160y-28z=-20 \\\\-153y+2z =-10 \end{matrix}\right$

$\left\{\begin{matrix} 7x+10y+40z=15 \\\\ -20x-160y-28z=-20 \\\\-2142y+28z =-140 \end{matrix}\right\,\,\,\,\Rightarrow\\\\\\ \left\{\begin{matrix} 7x+10y+40z=15 \\\\ -20x-160y-28z=-20 \\\\-2302y=-160 \end{matrix}\right$

etc.

Deve-se resolver o sistema linear acima, para concluir a solução.