Question

Ajutati-ma va rog. Nu stiu cum sa rezolv! Va rog!

Answer 1

$\displaystyle \mathtt{ \left\{\begin{array}{ccc}\mathtt{x+ay+2z=1}\\\mathtt{x+(2a-1)y+3z=1}\\\mathtt{x+ay+(a-3)z=1}\end{array}\right\Rightarrow A= \left(\begin{array}{ccc}\mathtt1&\mathtt{a}&\mathtt2\\\mathtt1&\mathtt{2a-1}&\mathtt3\\\mathtt1&\mathtt a&\mathtt{a-3}\end{array}\right)}\\ \\ \mathtt{a)~det(A)=a^2-6a+5}$

$\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt{a}&\mathtt2\\\mathtt1&\mathtt{2a-1}&\mathtt3\\\mathtt1&\mathtt a&\mathtt{a-3}\end{array}\right|=1 \cdot (2a-1) \cdot(a-3)+2\cdot1\cdot a+}\\ \\ \mathtt{+a\cdot3\cdot1-2\cdot(2a-1)\cdot1-a\cdot1\cdot(a-3)-1\cdot3\cdot a=}\\ \\ \mathtt{=2a^2-7a+3+2a+3a-4a+2-a^2+3a-3a=a^2-6a+5}$

$\displaystyle \mathtt{b)~det(A)=0}\\ \\ \mathtt{a^2-6a+5=0}\\ \\ \mathtt{a^2-a-5a+5=0}\\ \\ \mathtt{a(a-1)-5(a-1)=0}\\ \\ \mathtt{(a-1)(a-5)=0}\\ \\ \mathtt{a-1=0\Rightarrow a=1;~a-5=0\Rightarrow a=5}$

$\displaystyle \mathtt{c)~a=0\Rightarrow \left\{\begin{array}{ccc}\mathtt{x+2z=1}\\\mathtt{x-y+3z=1}\\\mathtt{x-3z=1}\end{array}\right\Rightarrow A= \left(\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt2\\\mathtt1&\mathtt{-1}&\mathtt3\\\mathtt1&\mathtt0&\mathtt{-3}\end{array}\right)}$

$\displaystyle \mathtt{\Delta=det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt2\\\mathtt1&\mathtt{-1}&\mathtt3\\\mathtt1&\mathtt0&\mathtt{-3}\end{array}\right|=1 \cdot(-1)\cdot(-3)+2\cdot1\cdot0+0\cdot3\cdot1-}\\ \\ \mathtt{-2\cdot(-1)\cdot1-0\cdot1\cdot(-3)-1\cdot3\cdot0=5}\\ \\ \mathtt{\Delta=det(A)=5\ne0}$

$\displaystyle \mathtt{\Delta_x=\left|\begin{array}{ccc}\mathtt1&\mathtt0&\mathtt2\\\mathtt1&\mathtt{-1}&\mathtt3\\\mathtt1&\mathtt0&\mathtt{-3}\end{array}\right|=1 \cdot(-1)\cdot(-3)+2\cdot1\cdot0+0\cdot3\cdot1-}\\ \\ \mathtt{-2\cdot(-1)\cdot1-0\cdot1\cdot(-3)-1\cdot3\cdot0=5}\\ \\ \mathtt{\Delta_x=5}$

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

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

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{5}{5}=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=1}\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta} = \frac{0}{5}=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=0 }~\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{0}{5}=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~z=0 }$