Question

sa se rezolve sistemul de ecuatii; x+y+z=6, 2x-y-z=7, x-y+2z=4. PAS CU PAS VA ROG!

Answer 1

$\displaystyle \mathtt{\left\{\begin{array}{ccc}\mathtt{x+y+z=6}\\\mathtt{2x-y-z=7}\\\mathtt{x-y+2z=4}\end{array}\right ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~A=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt1\\\mathtt2&\mathtt{-1}&\mathtt{-1}\\\mathtt1&\mathtt{-1}&\mathtt2\end{array}\right)}$

$\displaystyle \mathtt{\Delta=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt1\\\mathtt2&\mathtt{-1}&\mathtt{-1}\\\mathtt1&\mathtt{-1}&\mathtt2\end{array}\right|=1\cdot(-1)\cdot2+1\cdot2\cdot(-1)+1\cdot(-1)\cdot1-}\\ \\ \mathtt{-1\cdot(-1)\cdot1-1\cdot2\cdot2-1\cdot(-1)\cdot(-1)=-2-2-1+1-4-1=-9}\\ \\ \mathtt{\Delta=-9\ne0\Rightarrow Se~poate~aplica~Cramer}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt6&\mathtt1&\mathtt1\\\mathtt7&\mathtt{-1}&\mathtt{-1}\\\mathtt4&\mathtt{-1}&\mathtt2\end{array}\right|=6\cdot(-1)\cdot2+1\cdot7\cdot(-1)+1\cdot(-1)\cdot4-}\\ \\ \mathtt{-1\cdot(-1)\cdot4-1\cdot7\cdot2-6\cdot(-1)\cdot(-1)=}\\ \\ \mathtt{=-12-7-4+4-14-6=-39} \\ \\ \mathtt{\Delta_x=-39}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt1&\mathtt6&\mathtt1\\\mathtt2&\mathtt7&\mathtt{-1}\\\mathtt1&\mathtt4&\mathtt2\end{array}\right|=1 \cdot 7 \cdot2+1\cdot2\cdot4+6\cdot(-1)\cdot1-1\cdot7\cdot1-}\\ \\ \mathtt{-6\cdot2\cdot2-1\cdot(-1)\cdot4=14+8-6-7-24+4=-11}\\ \\ \mathtt{\Delta_y=-11}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt6\\\mathtt2&\mathtt{-1}&\mathtt7\\\mathtt1&\mathtt{-1}&\mathtt4\end{array}\right|=1\cdot(-1)\cdot4+6\cdot2\cdot(-1)+1\cdot7\cdot1-6\cdot(-1)\cdot1-}\\ \\ \mathtt{-1\cdot2\cdot4-1\cdot7\cdot(-1)=-4-12+7+6-8+7=-4}\\ \\ \mathtt{\Delta_z=-4}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta} = \frac{-39}{-9} = \frac{39}{9}= \frac{13}{3} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\mathtt{x= \frac{13}{3} }}}\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{-11}{-9}= \frac{11}{9} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\mathtt{y= \frac{11}{9} }}}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-4}{-9}= \frac{4}{9}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\mathtt{z= \frac{4}{9} }} }$