Matrizes Ajudem... Se A=![\left[\begin{array}{cc}2&1\\3&-1\end{array}\right] , B= \left[\begin{array}{cc}-1&2\\1&0\end{array} \right] C= \left[\begin{array}{cc}4&-1\\2&1\end{array}\right]](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26amp%3B1%5C%5C3%26amp%3B-1%5Cend%7Barray%7D%5Cright%5D++%2C+B%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26amp%3B2%5C%5C1%26amp%3B0%5Cend%7Barray%7D+%5Cright%5D+C%3D+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26amp%3B-1%5C%5C2%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+ "\left[\begin{array}{cc}2&1\\3&-1\end{array}\right] , B= \left[\begin{array}{cc}-1&2\\1&0\end{array} \right] C= \left[\begin{array}{cc}4&-1\\2&1\end{array}\right]")
determine a matriz X de ordem 2, tal que 
Soluções para a tarefa
Respondido por
13
Oi. Bom dia Izis
![\frac{X-A}{2} = \frac{B+X}{3} +C \\ \\ \frac{3(X-A)=2(B+X)+6C}{6} \\ \\ 3X-3A=2B+2X+6C \\ 3X-2X=3A+2B+6C \\ X=3. \left[\begin{array}{ccc}2&1\\3&-1\end{array}\right] +2. \left[\begin{array}{ccc}-1&2\\1&0\end{array}\right] +6. \left[\begin{array}{ccc}4&-1\\2&1\end{array}\right] \\ \\ X= \left[\begin{array}{ccc}6&3\\9&-3\end{array}\right] + \left[\begin{array}{ccc}-2&4\\2&0\end{array}\right] + \left[\begin{array}{ccc}24&-6\\12&6\end{array}\right]](https://tex.z-dn.net/?f=+%5Cfrac%7BX-A%7D%7B2%7D+%3D+%5Cfrac%7BB%2BX%7D%7B3%7D+%2BC+%5C%5C++%5C%5C++%5Cfrac%7B3%28X-A%29%3D2%28B%2BX%29%2B6C%7D%7B6%7D++%5C%5C++%5C%5C+3X-3A%3D2B%2B2X%2B6C+%5C%5C+3X-2X%3D3A%2B2B%2B6C+%5C%5C+X%3D3.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B1%5C%5C3%26amp%3B-1%5Cend%7Barray%7D%5Cright%5D+%2B2.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B2%5C%5C1%26amp%3B0%5Cend%7Barray%7D%5Cright%5D+%2B6.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26amp%3B-1%5C%5C2%26amp%3B1%5Cend%7Barray%7D%5Cright%5D++%5C%5C++%5C%5C+X%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%26amp%3B3%5C%5C9%26amp%3B-3%5Cend%7Barray%7D%5Cright%5D+%2B+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B4%5C%5C2%26amp%3B0%5Cend%7Barray%7D%5Cright%5D+%2B+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D24%26amp%3B-6%5C%5C12%26amp%3B6%5Cend%7Barray%7D%5Cright%5D++ "\frac{X-A}{2} = \frac{B+X}{3} +C \\ \\ \frac{3(X-A)=2(B+X)+6C}{6} \\ \\ 3X-3A=2B+2X+6C \\ 3X-2X=3A+2B+6C \\ X=3. \left[\begin{array}{ccc}2&1\\3&-1\end{array}\right] +2. \left[\begin{array}{ccc}-1&2\\1&0\end{array}\right] +6. \left[\begin{array}{ccc}4&-1\\2&1\end{array}\right] \\ \\ X= \left[\begin{array}{ccc}6&3\\9&-3\end{array}\right] + \left[\begin{array}{ccc}-2&4\\2&0\end{array}\right] + \left[\begin{array}{ccc}24&-6\\12&6\end{array}\right]")
![X= \left[\begin{array}{ccc}28&1\\23&3\end{array}\right]](https://tex.z-dn.net/?f=X%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D28%26amp%3B1%5C%5C23%26amp%3B3%5Cend%7Barray%7D%5Cright%5D "X= \left[\begin{array}{ccc}28&1\\23&3\end{array}\right]")
Perguntas interessantes