Question

la punctul a de la E8 dau coroană pentru cel mai bun răspuns. va rog!! urgent!!!!

Answer 1

$\displaystyle \mathtt{X \cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt2\\\mathtt3&\mathtt4\end{array}\right)=\left(\begin{array}{ccc}\mathtt5&\mathtt{10}\\\mathtt4&\mathtt2\end{array}\right)}\\ \\ \mathtt{XA=B\Rightarrow X=BA^{-1}}\\ \\ \mathtt{A=\left(\begin{array}{ccc}\mathtt{-1}&\mathtt2\\\mathtt3&\mathtt4\end{array}\right);~B=\left(\begin{array}{ccc}\mathtt5&\mathtt{10}\\\mathtt4&\mathtt2\end{array}\right)}$

$\displaystyle \mathtt{\displaystyle \mathtt{X \cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt2\\\mathtt3&\mathtt4\end{array}\right)=\left(\begin{array}{ccc}\mathtt5&\mathtt{10}\\\mathtt4&\mathtt2\end{array}\right)}} \\ \\ \mathtt{A^{-1}= \frac{1}{det(A)} \cdot A^*}$

$\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt2\\\mathtt3&\mathtt4\end{array}\right| =(-1)\cdot4-2 \cdot 3=-4-6=-10\ne 0}$

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt{-1}&\mathtt2\\\mathtt3&\mathtt4\end{array}\right)\Rightarrow A^{tr}=\left(\begin{array}{ccc}\mathtt{-1}&\mathtt3\\\mathtt2&\mathtt4\end{array}\right)}$

$\displaystyle \mathtt{A^*=\left(\begin{array}{ccc}\mathtt{(-1)^{1+1}\cdot4}&\mathtt{(-1)^{1+2}\cdot 2}\\\mathtt{(-1)^{2+1}\cdot3}&\mathtt{(-1)^{2+2}\cdot (-1)}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{1\cdot4}&\mathtt{(-1)\cdot2}\\\mathtt{(-1)\cdot 3}&\mathtt{1\cdot(-1)}\end{array}\right)=}\\ \\ \\\mathtt{=\left(\begin{array}{ccc}\mathtt{4}&\mathtt{-2}\\\mathtt{-3}&\mathtt{-1}\end{array}\right)}$

$\displaystyle\mathtt{A^{-1}=-\frac{1}{10}\cdot\left(\begin{array}{ccc}\mathtt{4}&\mathtt{-2}\\\mathtt{-3}&\mathtt{-1}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{\left(\displaystyle-\frac{1}{10}\right)\cdot 4 }&\mathtt{\left(\displaystyle-\frac{1}{10}\right)\cdot(-2) }\\\\\mathtt{\displaystyle \left(-\frac{1}{10}\right)\cdot(-3) }&\mathtt{\displaystyle\left(-\frac{1}{10}\right)\cdot(-1)}\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{2}{5} }&\mathtt{\displaystyle \frac{1}{5} }\\\\\mathtt{\displaystyle \frac{3}{10} }&\mathtt{ \displaystyle\frac{1}{10} }\end{array}\right)}$

$\displaystyle \mathtt{X=\left(\begin{array}{ccc}\mathtt5&\mathtt{10}\\\mathtt4&\mathtt2\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{2}{5} }&\mathtt{\displaystyle \frac{1}{5} }\\\\\mathtt{\displaystyle \frac{3}{10} }&\mathtt{ \displaystyle\frac{1}{10} }\end{array}\right)}$

$\displaystyle \mathtt{X=\left(\begin{array}{ccc}\mathtt{\displaystyle 5\cdot\left( -\frac{2}{5}\right)+10\cdot \frac{3}{10}}&\mathtt{\displaystyle 5\cdot \frac{1}{5}+10\cdot \frac{1}{10} }\\\\\mathtt{\displaystyle 4\cdot\left(- \frac{2}{5}\right)+2\cdot \frac{3}{10} }&\mathtt{ \displaystyle 4 \cdot \frac{1}{5}+2 \cdot \frac{1}{10} }\end{array}\right)}$

$\displaystyle \mathtt{X=\left(\begin{array}{ccc}\mathtt{\displaystyle (-2)+3 }&\mathtt{\displaystyle 1+1 }\\\\\mathtt{\displaystyle \left(- \frac{8}{5}\right)+ \frac{3}{5} }&\mathtt{ \displaystyle \frac{4}{5}+ \frac{1}{5} }\end{array}\right)}\\ \\ \\ \mathtt{X= \left(\begin{array}{ccc}\mathtt{1}&\mathtt2\\\mathtt{-1}&\mathtt1\\\end{array}\right)}$