Question

Am nevoie de răspunsul la b și c

Answer 1

$\displaystyle \mathtt{ \left\{\begin{array}{ccc}\mathtt{-x+ay+(2a+4)y=1}\\\mathtt{(a+2)x+ay+(a+1)z=1}\\\mathtt{(a+1)x+(2a-1)y+3z=2}\end{array}\right,~a \in\mathbb{R}}$

$\displaystyle \mathtt{b)~Sistemul~este~compatibil~determinat\Leftrightarrow\Delta\ne0}\\ \\ \mathtt{3a^3+9a^2-3a-9\ne 0}\\ \\ \mathtt{3a^3+9a^2-3a-9=0} \\ \\ \mathtt{3a^2(a+3)-3(a+3)=0}\\ \\ \mathtt{(a+3)\left(3a^2-3\right)=0}\\ \\ \mathtt{(a+3)3\left(a^2-1\right)=0}\\ \\ \mathtt{3(a+3)(a+1)(a-1)= 0}\\ \\ \mathtt{a=-3;~a=-1;~a=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a\in\mathbb{R}-\{-3;-1;1\}}$

$\displaystyle \mathtt{c)~a=-2\Rightarrow \left\{\begin{array}{ccc}\mathtt{-x-2y=1}\\\mathtt{-2y-z=1}\\\mathtt{-x-5y+3z=2}\end{array}\right\Rightarrow A= \left(\begin{array}{ccc}\mathtt{-1}&\mathtt{-2}&\mathtt0\\\mathtt0&\mathtt{-2}&\mathtt{-1}\\\mathtt{-1}&\mathtt{-5}&\mathtt3\end{array}\right)}$

$\displaystyle \mathtt{\Delta=det(A)=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt{-2}&\mathtt0\\\mathtt0&\mathtt{-2}&\mathtt{-1}\\\mathtt{-1}&\mathtt{-5}&\mathtt3\end{array}\right|=(-1)\cdot(-2)\cdot3+0\cdot0\cdot(-5)+}\\ \\ \mathtt{+(-2)\cdot(-1)\cdot(-1)-0\cdot(-2)\cdot(-1)-(-2)\cdot0\cdot3-(-1)\cdot(-1)\cdot(-5)=}\\ \\ \mathtt{=6-2+5=9}\\ \\ \mathtt{\Delta=det(A)=9\ne0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt1&\mathtt{-2}&\mathtt0\\\mathtt1&\mathtt{-2}&\mathtt{-1}\\\mathtt2&\mathtt{-5}&\mathtt3\end{array}\right|=1\cdot(-2)\cdot3+0\cdot1\cdot(-5)+(-2)\cdot(-1)\cdot2-}\\ \\ \mathtt{-0\cdot(-2)\cdot2-(-2)\cdot1\cdot3-1\cdot(-1)\cdot(-5)=-6+4+6-5=-1}\\ \\ \mathtt{\Delta_x=-1}$

$\displaystyle \mathtt{\Delta_y=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt0\\\mathtt0&\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt2&\mathtt3\end{array}\right|=(-1)\cdot1\cdot3+0\cdot0\cdot2+1\cdot(-1)\cdot(-1)-}\\ \\ \mathtt{-0\cdot1\cdot(-1)-1\cdot0\cdot3-(-1)\cdot(-1)\cdot2=-3+1-2=-4}\\ \\ \mathtt{\Delta_y=-4}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt{-2}&\mathtt1\\\mathtt{-1}&\mathtt{-5}&\mathtt2\end{array}\right|=(-1)\cdot(-2)\cdot2+1\cdot0\cdot(-5)+(-2)\cdot1\cdot(-1)-}\\ \\ \mathtt{-1\cdot(-2)\cdot(-1)-(-2)\cdot0\cdot2-(-1)\cdot1\cdot(-5)=4+2-2-5=-1}\\ \\ \mathtt{\Delta_z=-1}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{-1}{9}=- \frac{1}{9}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=- \frac{1}{9} }\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{-4}{9}=- \frac{4}{9}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=- \frac{4}{9} }\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-1}{9}=- \frac{1}{9} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z=- \frac{1}{9} }$