Question

Va rog cn ma ajuta si pe mine..sunt la lucrare, cineva care sa fie bun la matematica:sa se rezolve prin metoda Cramer sistemul:
2x+3y+4z=-2
-x+2y+3z=5
3x-y-z=-11

Answer 1

sper sa se înțeleagă.mult succes

Answer 2

$\displaystyle \mathtt{ \left\{\begin{array}{ccc}\mathtt{2x+3y+4z=-2}\\\mathtt{-x+2y+3z=5}\\\mathtt{3x-y-z=-11}\end{array}\right \Rightarrow A= \left(\begin{array}{ccc}\mathtt2&\mathtt3&\mathtt4\\\mathtt{-1}&\mathtt2&\mathtt3\\\mathtt3&\mathtt{-1}&\mathtt{-1}\end{array}\right)}$

$\displaystyle \mathtt{ \Delta=det(A)=\left|\begin{array}{ccc}\mathtt2&\mathtt3&\mathtt4\\\mathtt{-1}&\mathtt2&\mathtt3\\\mathtt3&\mathtt{-1}&\mathtt{-1}\end{array}\right|=2 \cdot 2 \cdot(-1)+4\cdot(-1)\cdot(-1)+}\\ \\ \mathtt{+3 \cdot3\cdot3-4\cdot2\cdot3-3\cdot(-1)\cdot(-1)-2\cdot3\cdot(-1)=}\\ \\ \mathtt{=-4+4+27-24-3+6=6}\\ \\ \mathtt{\Delta=det(A)=6 \ne 0}$

$\displaystyle \mathtt{\Delta_x= \left|\begin{array}{ccc}\mathtt{-2}&\mathtt3&\mathtt4\\\mathtt5&\mathtt2&\mathtt3\\\mathtt{-11}&\mathtt{-1}&\mathtt{-1}\end{array}\right|=(-2)\cdot2\cdot(-1)+4\cdot5\cdot(-1)+3\cdot3\cdot(-11)-}\\ \\ \mathtt{-4\cdot2\cdot(-11)-3\cdot5\cdot(-1)-(-2)\cdot3\cdot(-1)=}\\ \\ \mathtt{=4-20-99+88+15-6=-18}\\ \\ \mathtt{\Delta_x=-18}$

$\displaystyle \mathtt{\Delta_y= \left|\begin{array}{ccc}\mathtt2&\mathtt{-2}&\mathtt4\\\mathtt{-1}&\mathtt5&\mathtt3\\\mathtt3&\mathtt{-11}&\mathtt{-1}\end{array}\right|=2\cdot5\cdot(-1)+4\cdot(-1)\cdot(-11)+(-2)\cdot3\cdot3-}\\ \\ \mathtt{-4\cdot5\cdot3-(-2)\cdot(-1)\cdot(-1)-2\cdot3\cdot(-11)=}\\ \\ \mathtt{=-10+44-18-60+2+66=24}\\ \\ \mathtt{\Delta_y=24}$

$\displaystyle \mathtt{\Delta_z= \left|\begin{array}{ccc}\mathtt2&\mathtt3&\mathtt{-2}\\\mathtt{-1}&\mathtt2&\mathtt5\\\mathtt3&\mathtt{-1}&\mathtt{-11}\end{array}\right|=2\cdot2\cdot(-11)+(-2)\cdot(-1)\cdot(-1)+3\cdot5\cdot3-}\\ \\ \mathtt{-(-2)\cdot2\cdot3-3\cdot(-1)\cdot(-11)-2\cdot5\cdot(-1)=}\\ \\ \mathtt{=-44-2+45+12-33+10=-12}\\ \\ \mathtt{\Delta_z=-12}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{-18}{6}= -3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=-3 }\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta} = \frac{24}{6}= 4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=4}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-12}{6}=-2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z=-2}$