Question

Urgent va rooog frumos

Answer 1

[tex]\displaystyle \mathtt{\left\{\begin{array}{ccc}\mathtt{x+y-2z=0}\\\mathtt{x-y+z=1}\\\mathtt{x+y+az=2}\end{array}\right,~a\in\mathbb{R}~~~~~~~~~~~~~~~~~~~~~~~~A= \left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-2}\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt1&\mathtt1&\mathtt a\end{array}\right)}[/tex]

[tex]\displaystyle \mathtt{a)~det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-2}\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt1&\mathtt1&\mathtt a\end{array}\right|=1 \cdot (-1) \cdot a+(-2)\cdot1\cdot1+1\cdot1\cdot1-}\\ \\ \mathtt{-(-2)\cdot(-1)\cdot1-1\cdot1\cdot a-1\cdot1\cdot1=-a-2+1-2-a-1=}\\ \\ \mathtt{=-2a-4}}[/tex]

[tex]\displaystyle \mathtt{b)~Matricea~A~este~inversabil\u{a}~\Leftrightarrow~det(A)\ne 0}\\ \\ \mathtt{det(A)=-2a-4}\\ \\ \mathtt{-2a-4\ne0 \Rightarrow a\ne -2}\\ \\ \mathtt{Pentru~a\in\mathbb{R}-\{-2\}~matricea~A~este~inversabil\u{a}~.}[/tex]

[tex]\displaystyle \mathtt{c)~a=0\Rightarrow \mathtt{\left\{\begin{array}{ccc}\mathtt{x+y-2z=0}\\\mathtt{x-y+z=1}\\\mathtt{x+y=2}\end{array}\right}}[/tex]

[tex]\displaystyle \mathtt{\Delta= \left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-2}\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt1&\mathtt1&\mathtt 0\end{array}\right|=1 \cdot (-1) \cdot0+(-2)\cdot 1 \cdot 1+1\cdot1\cdot1-}\\ \\ \mathtt{-(-2)\cdot(-1)\cdot1-1 \cdot 1 \cdot0-1\cdot1\cdot1=-4}\\ \\ \mathtt{\Delta=-4\ne0}[/tex]

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

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

$\displaystyle \mathtt{\Delta_z= \left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt1&\mathtt{-1}&\mathtt1\\\mathtt1&\mathtt1&\mathtt 2\end{array}\right|=1 \cdot (-1) \cdot 2+0 \cdot 1 \cdot 1+1 \cdot 1 \cdot 1-0 \cdot (-1) \cdot 1-}\\ \\ \mathtt{-1 \cdot 1 \cdot 2-1 \cdot 1 \cdot 1=-4}\\ \\ \mathtt{\Delta_z=-4}$

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