como resolver essa questão, se alguém que saiba ajudar, agradeço
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Olá!
![\left[\begin{array}{lcr}-7&2&1\\6&4&-3\end{array}\right]+2X = \left[\begin{array}{lcr}11&0&3\\ 8&12&5\end{array}\right]\Rightarrow \\ \\ \\ \Rightarrow
2X = \left[\begin{array}{lcr}11&0&3\\ 8&12&5\end{array}\right]- \left[\begin{array}{lcr}-7&2&1\\6&4&-3\end{array}\right] = \\ \\ \\ =
\left[\begin{array}{lcr}(11+7)&(0-2)&(3-1)\\ (8-6)& (12-4)& (5+3) \end{array}\right]=
\left[\begin{array}{lcr}18&-2&2\\ 2&8&8 \end{array}\right].](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D-7%26amp%3B2%26amp%3B1%5C%5C6%26amp%3B4%26amp%3B-3%5Cend%7Barray%7D%5Cright%5D%2B2X+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D11%26amp%3B0%26amp%3B3%5C%5C+8%26amp%3B12%26amp%3B5%5Cend%7Barray%7D%5Cright%5D%5CRightarrow+%5C%5C+%5C%5C+%5C%5C+%5CRightarrow+%0A2X+%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D11%26amp%3B0%26amp%3B3%5C%5C+8%26amp%3B12%26amp%3B5%5Cend%7Barray%7D%5Cright%5D-+%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D-7%26amp%3B2%26amp%3B1%5C%5C6%26amp%3B4%26amp%3B-3%5Cend%7Barray%7D%5Cright%5D+%3D+%5C%5C+%5C%5C+%5C%5C+%3D+%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D%2811%2B7%29%26amp%3B%280-2%29%26amp%3B%283-1%29%5C%5C+%288-6%29%26amp%3B+%2812-4%29%26amp%3B+%285%2B3%29+%5Cend%7Barray%7D%5Cright%5D%3D%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D18%26amp%3B-2%26amp%3B2%5C%5C+2%26amp%3B8%26amp%3B8+%5Cend%7Barray%7D%5Cright%5D. "\left[\begin{array}{lcr}-7&2&1\\6&4&-3\end{array}\right]+2X = \left[\begin{array}{lcr}11&0&3\\ 8&12&5\end{array}\right]\Rightarrow \\ \\ \\ \Rightarrow
2X = \left[\begin{array}{lcr}11&0&3\\ 8&12&5\end{array}\right]- \left[\begin{array}{lcr}-7&2&1\\6&4&-3\end{array}\right] = \\ \\ \\ =
\left[\begin{array}{lcr}(11+7)&(0-2)&(3-1)\\ (8-6)& (12-4)& (5+3) \end{array}\right]=
\left[\begin{array}{lcr}18&-2&2\\ 2&8&8 \end{array}\right].")
Daí,
![X=\dfrac{1}{2} \left[\begin{array}{lcr}18&-2&2\\ 2&8&8 \end{array}\right] =
\left[\begin{array}{lcr}(\frac{1}{2}\cdot 18)&(\frac{1}{2}\cdot (-2))&(\frac{1}{2}\cdot 2)\\ \; & \; & \; \\ (\frac{1}{2}\cdot 2) & (\frac{1}{2}\cdot 8)&(\frac{1}{2}\cdot 8)\end{array}\right]\Rightarrow\\ \\ \\ \Rightarrow
X=\left[\begin{array}{lcr}9&-1&1\\ 1&4&4\end{array}\right].](https://tex.z-dn.net/?f=X%3D%5Cdfrac%7B1%7D%7B2%7D++%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D18%26amp%3B-2%26amp%3B2%5C%5C+2%26amp%3B8%26amp%3B8+%5Cend%7Barray%7D%5Cright%5D+%3D+%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot+18%29%26amp%3B%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot+%28-2%29%29%26amp%3B%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot+2%29%5C%5C+%5C%3B+%26amp%3B+%5C%3B+%26amp%3B+%5C%3B+%5C%5C+%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot+2%29+%26amp%3B+%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot+8%29%26amp%3B%28%5Cfrac%7B1%7D%7B2%7D%5Ccdot+8%29%5Cend%7Barray%7D%5Cright%5D%5CRightarrow%5C%5C+%5C%5C+%5C%5C+%5CRightarrow%0AX%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Blcr%7D9%26amp%3B-1%26amp%3B1%5C%5C+1%26amp%3B4%26amp%3B4%5Cend%7Barray%7D%5Cright%5D.%0A "X=\dfrac{1}{2} \left[\begin{array}{lcr}18&-2&2\\ 2&8&8 \end{array}\right] =
\left[\begin{array}{lcr}(\frac{1}{2}\cdot 18)&(\frac{1}{2}\cdot (-2))&(\frac{1}{2}\cdot 2)\\ \; & \; & \; \\ (\frac{1}{2}\cdot 2) & (\frac{1}{2}\cdot 8)&(\frac{1}{2}\cdot 8)\end{array}\right]\Rightarrow\\ \\ \\ \Rightarrow
X=\left[\begin{array}{lcr}9&-1&1\\ 1&4&4\end{array}\right].")
Bons estudos!
Daí,
Bons estudos!
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