Question

Ajutor va rog, am nevoie de ajutor pentru aceste 3 exerciti trebuie sa le fac pana maine si nu am idee cum, va rog mult sa ma ajutați!!!

Answer 1

$\displaystyle \mathtt{41.~ \left\{\begin{array}{ccc}\mathtt{x+y+az=1}\\\mathtt{x+2ay+z=-1}\\\mathtt{2ax+y+(a+1)z}=0\end{array}\right,x,y,z\in\mathbb{R}~~~~a\in\mathbb{R}}$

$\displaystyle \mathtt{a)~a=0\Rightarrow \left\{\begin{array}{ccc}\mathtt{x+y=1}\\\mathtt{x+z=-1}\\\mathtt{y+z}=0\end{array}\Rightarrow A= \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt1&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{\Delta=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt1&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\end{array}\right|=1 \cdot 0 \cdot 1+0 \cdot 1\cdot1+1 \cdot 1 \cdot 0-0\cdot0\cdot0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1 \cdot 1 \cdot 1-1 \cdot 1 \cdot 1=-2}\\ \\ \mathtt{\Delta =-2\neq0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt{-1}&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt1\end{array}\right|=1 \cdot0\cdot1+0 \cdot (-1)\cdot1+1\cdot1\cdot0-0\cdot0\cdot0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1 \cdot (-1)\cdot1-1 \cdot 1 \cdot 1=0}\\ \\ \mathtt{\Delta_x=0}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt0&\mathtt1\end{array}\right|=1 \cdot (-1)\cdot1+0\cdot1\cdot0+1\cdot1\cdot0-0\cdot(-1)\cdot0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1\cdot1\cdot1-1\cdot1\cdot0=-2} \\ \\ \mathtt{\Delta_y=-2}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt1\\\mathtt1&\mathtt0&\mathtt{-1}\\\mathtt0&\mathtt1&\mathtt0\end{array}\right|=1 \cdot 0 \cdot 0+1 \cdot 1 \cdot1+1 \cdot(-1)\cdot0-1\cdot0\cdot0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1 \cdot 1 \cdot 0-1 \cdot (-1)\cdot1=2}\\ \\ \mathtt{\Delta_z=2}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{0}{-2}=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=0}\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{-2}{-2} =1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=1}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta} = \frac{2}{-2}=-1~~~~~~~~~~~~~~~~~~~~~~~~~~~z=-1}$

$\displaystyle \mathtt{b)~a=-1\Rightarrow\left\{\begin{array}{ccc}\mathtt{x+y-z=1}\\\mathtt{x-2y+z=-1}\\\mathtt{-2x+y=0}\end{array}\right}\mathtt{\Rightarrow A= \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt{-2}&\mathtt1&\mathtt0\end{array}\right)}$

$\mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt{-2}&\mathtt1&\mathtt0\end{array}\right|=1\cdot(-2)\cdot0+(-1)\cdot1\cdot1+1 \cdot1\cdot(-2)-}\\\\\mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-(-1)\cdot (-2)\cdot(-2)-1\cdot1\cdot0-1\cdot1\cdot1=0}$

$\mathtt{^A~\widetilde{~}\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt{-2}&\mathtt1\\\mathtt{-2}&\mathtt1&\mathtt0\end{array}\right)\widetilde{}\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt{-3}&\mathtt2\\\mathtt{-2}&\mathtt1&\mathtt0\end{array}\right)\widetilde{~} \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt{-3}&\mathtt2\\\mathtt{0}&\mathtt3&\mathtt{-2}\end{array}\right)\widetilde{~}\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt{-3}&\mathtt2\\\mathtt{0}&\mathtt0&\mathtt0\end{array}\right)}$

$\mathtt{rang(A)=2}\\ \\ \mathtt{rang(\overline{A})=2}\\ \\ \mathtt{Sistemul~este~compatibil.}$

$\displaystyle\mathtt{c)~\Delta\neq 0}\\\\\mathtt{\Delta=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt a\\\mathtt1&\mathtt{2a}&\mathtt1\\\mathtt{2a}&\mathtt1&\mathtt{a+1}\end{array}\right|=1\cdot 2a\cdot(a+1)+a\cdot 1\cdot1+1\cdot1\cdot2a-}\\\\ \mathtt{~~~~~~~~~~~~~-a\cdot2a\cdot2a-1\cdot1\cdot (a+1)-1\cdot1\cdot1=-4a^3+2a^2+4a-2}\\\\\mathtt{-4a^3+2a^2+4a-2\neq0}\\\\ \mathtt{-2(a-1)(a+1)(2a-1)\neq0}\\\\\mathtt{a\neq1;~a\neq-1;~a\neq \frac{1}{2}}\\\\\mathtt{m\in\mathbb{R}-\left\{-1,\frac{1}{2},1\right\}}$