Question

x+y+z=4, x-y+z=2, -x+y+z=2 ,x,y,z=?

Answer 1

aduni prima cu a doua si vine 2x+2z=6 x+z=3
prima cu a treia: y+z=3
a doua cu a treia z=2
y=1 x=1

Answer 2

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

$\displaystyle \mathtt{\Delta=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt{-1}&\mathtt1&\mathtt1\end{array}\right|=1 \cdot (-1)\cdot 1+1 \cdot1\cdot1+1\cdot1\cdot(-1)-}\\ \\ \mathtt{-1\cdot(-1)\cdot(-1)-1 \cdot 1\cdot1-1\cdot1\cdot1=-1+1-1-1-1-1=-4}\\ \\ \mathtt{\Delta=-4\ne 0}$

$\displaystyle \mathtt{\Delta=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt{-1}&\mathtt1&\mathtt1\end{array}\right|=1 \cdot (-1)\cdot 1+1 \cdot1\cdot1+1\cdot1\cdot(-1)-}\\ \\ \mathtt{-1\cdot(-1)\cdot(-1)-1 \cdot 1\cdot1-1\cdot1\cdot1=-1+1-1-1-1-1=-4}\\ \\ \mathtt{\Delta=-4\ne 0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt4&\mathtt1&\mathtt1\\\mathtt2&\mathtt{-1}&\mathtt1\\\mathtt2&\mathtt1&\mathtt1\end{array}\right|=4 \cdot (-1)\cdot1+1 \cdot2\cdot1+1\cdot1\cdot2-1\cdot(-1)\cdot2-}\\ \\ \mathtt{-1\cdot2\cdot1-4\cdot1\cdot1=-4+2+2+2-2-4=-4}\\ \\ \mathtt{\Delta_x=-4}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt1&\mathtt4&\mathtt1\\\mathtt1&\mathtt2&\mathtt1\\\mathtt{-1}&\mathtt2&\mathtt1\end{array}\right|=1\cdot2\cdot1+1 \cdot 1\cdot2+4\cdot1\cdot(-1)-1\cdot2\cdot(-1)-} \\ \\ \mathtt{-4\cdot1\cdot1-1\cdot1\cdot2=2+2-4+2-4-2=-4}\\ \\ \mathtt{\Delta_y=-4}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt4\\\mathtt1&\mathtt{-1}&\mathtt2\\\mathtt{-1}&\mathtt1&\mathtt2\end{array}\right|=1\cdot(-1)\cdot2+4 \cdot 1 \cdot 1+1 \cdot 2 \cdot (-1)-}\\ \\ \mathtt{-4 \cdot (-1)\cdot(-1)-1 \cdot 1\cdot2-1\cdot2\cdot1=-2+4-2-4-2-2=-8}\\ \\ \mathtt{\Delta_z=-8}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{-4}{-4}=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\mathtt{x=1}} }\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{-4}{-4}=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\mathtt{y=1}}}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-8}{-4}=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\mathtt{z=2}} }$