Question

sa se rezolve sistemul

Answer 1

$\displaystyle \mathtt{ \left\{\begin{array}{ccc}\mathtt{-x+y+z=-2}\\\mathtt{5x-2y+z=-2}\\\mathtt{2y+3z=-2}\end{array}\right \Rightarrow A= \left(\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\maathtt0&\mathtt2&\mathtt3\end{array}\right)}$

$\displaystyle \mathtt{\Delta=det(A)=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\maathtt0&\mathtt2&\mathtt3\end{array}\right|=(-1)\cdot(-2)\cdot3+1\cdot5\cdot2+1\cdot1\cdot0-}\\ \\ \mathtt{-1\cdot(-2)\cdot0-1\cdot5\cdot3-(-1)\cdot1\cdot2=6+10-15+2=3}\\ \\ \mathtt{\Delta=det(A)=3\ne0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt{-2}&\mathtt1&\mathtt1\\\mathtt{-2}&\mathtt{-2}&\mathtt1\\\maathtt{-2}&\mathtt2&\mathtt3\end{array}\right|=(-2)\cdot(-2)\cdot3+1\cdot(-2)\cdot2+1\cdot1\cdot(-2)-}\\ \\ \mathtt{-1\cdot(-2)\cdot(-2)-1\cdot(-2)\cdot3-(-2)\cdot1\cdot2=}\\ \\ \mathtt{=12-4-2-4+6+4=12}\\ \\ \mathtt{\Delta_x=12}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt{-2}&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\maathtt0&\mathtt{-2}&\mathtt3\end{array}\right|=(-1)\cdot(-2)\cdot3+1\cdot5\cdot(-2)+(-2)\cdot1\cdot0-}\\ \\ \mathtt{-1\cdot(-2)\cdot0-(-2)\cdot5\cdot3-(-1)\cdot1\cdot(-2)=6-10+30-2=24}\\ \\ \mathtt{\Delta_y=24}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt{-2}\\\mathtt5&\mathtt{-2}&\mathtt{-2}\\\maathtt0&\mathtt2&\mathtt{-2}\end{array}\right|=(-1)\cdot(-2)\cdot(-2)+(-2)\cdot5\cdot2+1\cdot(-2)\cdot0-} \\ \\ \mathtt{-(-2)\cdot(-2)\cdot0-1\cdot5\cdot(-2)-(-1)\cdot(-2)\cdot2=}\\ \\ \mathtt{=-4-20+10-4=-18}\\ \\ \mathtt{\Delta_z=-18}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta} = \frac{12}{3} =4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=4}\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{24}{3}=8~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=8}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-18}{3}=-6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z=-6 }$