Question

Ştie cineva cum se face la 1 ? a, b, c ? Va rog

Answer 1

Sper că ai înțeles ce am făcut. Dacă nu înțelegi mă poți întreba.

Answer 2

$\displaystyle \mathtt{1.~} \left\{\begin{array}{ccc}\mathtt{x-3y+az=1}\\\mathtt{x+2y+z=2}\\\mathtt{ax+y-z=3}\end{array}\right\\ \\ \mathtt{a)~x=1,~y=2,~z=-3}\\ \\ \mathtt{\left\{ {1-3 \cdot 2+a \cdot (-3)=1} \atop {a \cdot 1+2-(-3)=3}} \right.\Rightarrow \left \{ {{1-6-3a=1} \atop {a+2+3=3}} \right. \Rightarrow \left \{ {{-3a=6}\atop {a=-2}} \right. }\\ \\ \mathtt{a=-2}$

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

$\displaystyle \mathtt{det(A)= \left|\begin{array}{ccc}\mathtt1&\mathtt{-3}&\mathtt2\\\mathtt1&\mathtt2&\mathtt1\\\mathtt2&\mathtt1&\mathtt{-1}\end{array}\right|=1 \cdot 2 \cdot (-1)+2 \cdot 1 \cdot 1+(-3) \cdot 1\cdot2-}\\ \\ \mathtt{-2 \cdot 2 \cdot 2-(-3) \cdot 1 \cdot(-1)-1\cdot1\cdot1=-2+2-6-8-3-1=-18}\\ \\ \mathtt{det(A)=-18}$

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

$\displaystyle \mathtt{\Delta=det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt{-3}&\mathtt{-2}\\\mathtt1&\mathtt2&\mathtt1\\\mathtt{-2}&\mathtt1&\mathtt{-1}\end{array}\right|=1 \cdot 2 \cdot (-1)+(-2)\cdot1\cdot1+}\\ \\ \mathtt{+(-3)\cdot1\cdot(-2)-(-2)\cdot2\cdot(-2)-(-3)\cdot1\cdot(-1)-1\cdot1\cdot1=}\\ \\ \mathtt{=-2-2+6-8-3-1=-10}\\ \\ \mathtt{\Delta=det(A)=-10\ne0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt1&\mathtt{-3}&\mathtt{-2}\\\mathtt2&\mathtt2&\mathtt1\\\mathtt3&\mathtt1&\mathtt{-1}\end{array}\right|=1 \cdot 2 \cdot (-1)+(-2)\cdot2\cdot1+(-3)\cdot1\cdot3-} \\ \\ \mathtt{-(-2)\cdot 2 \cdot 3-(-3)\cdot2\cdot(-1)-1\cdot1\cdot1=-2-4-9+12-6-1=-10}\\ \\ \mathtt{\Delta_x=-10}$

$\mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-2}\\\mathtt1&\mathtt2&\mathtt1\\\mathtt{-2}&\mathtt3&\mathtt{-1}\end{array}\right|=1 \cdot 2 \cdot (-1)+(-2)\cdot1\cdot3+1\cdot1\cdot(-2)-}\\ \\ \mathtt{-(-2)\cdot2\cdot(-2)-1\cdot1\cdot(-1)-1\cdot1\cdot3=-2-6-2-8+1-3=-20} \\ \\ \mathtt{\Delta_y=-20}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt1&\mathtt{-3}&\mathtt1\\\mathtt1&\mathtt2&\mathtt2\\\mathtt{-2}&\mathtt1&\mathtt3\end{array}\right|=1 \cdot 2 \cdot 3+1\cdot1\cdot1+(-3)\cdot2\cdot (-2)-}\\ \\ \mathtt{-1 \cdot 2\cdot(-2)-(-3)\cdot1\cdot3-1\cdot2\cdot1=6+1+12+4+9-2=30} \\ \\ \mathtt{\Delta_z=30}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{-10}{-10} =1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=1 }\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta} = \frac{-20}{-10} =2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=2}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{30}{-10}=-3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z=-3}$