Question

subiectul 2 , exercitiul 1 , din poza .. va rog mult

Answer 1

$\displaystyle \left\{\begin{array}{ccc}x+y-2z=0\\x-y+z=1\\x+y+az=2\end{array}\right ,a \in R \\ \\ a)A= \left(\begin{array}{ccc}1&1&-2\\1&-1&1\\1&1&a\end{array}\right)$
$\displaystyle det(A)= \left|\begin{array}{ccc}1&1&-2\\1&-1&1\\1&1&a\end{array}\right|=\\ \\ =1 \cdot (-1) \cdot a+(-2) \cdot 1 \cdot 1+1 \cdot 1 \cdot 1-(-2) \cdot (-1) \cdot 1-1 \cdot 1 \cdot a-1 \cdot 1 \cdot 1 \\ \\ =-a-2+1-2-a-1=-2a-4$

$\displaystyle b)Matricea~asociata~sistemului~este~inversabila \Leftrightarrow det(A) \not =0\\ \\ det(A)=-2a-4 \\ \\ det(A) \not =0 \Rightarrow -2a-4 \not = 0 \Rightarrow -2a \not = 4 \Rightarrow a \not = -2 \\ \\ a\in R- \{-2\}$
$\displaystyle c)a=0\\ \\ \left\{\begin{array}{ccc}x+y-2z=0\\x-y+z=1\\x+y=2\end{array}\right \\ \\ \Delta= \left|\begin{array}{ccc}1&1&-2\\1&-1&1\\1&1&0\end{array}\right|= \\ \\ =1 \cdot (-1) \cdot 0+(-2) \cdot 1 \cdot 1+1 \cdot 1 \cdot 1-(-2) \cdot (-1) \cdot 1-1 \cdot 1 \cdot 0-1 \cdot 1\cdot 1\\ \\ =0-2+1-2-0-1=-4$
$\displaystyle \Delta_x= \left|\begin{array}{ccc}0&1&-2\\1&-1&1\\2&1&0\end{array}\right|= \\ \\ =0 \cdot (-1) \cdot0 +(-2) \cdot 1 \cdot 1+1 \cdot 1 \cdot 2-(-2) \cdot (-1) \cdot 2-1 \cdot 1 \cdot 0-0 \cdot 1 \cdot 1= \\ \\ =0-2+2-4-0-0=-4\\ \\ x= \frac{\Delta_x}{\Delta} = \frac{-4}{-4} =1$
$\displaystyle \Delta_y= \left|\begin{array}{ccc}1&0&-2\\1&1&1\\1&2&0\end{array}\right|= \\ \\ =1 \cdot 1 \cdot 0+(-2) \cdot 1 \cdot 2+0 \cdot 1 \cdot 1-(-2) \cdot 1 \cdot 1-0 \cdot 1\cdot 0-1 \cdot 1 \cdot 2=\\ \\ =0-4+0+2-0-2=-4 \\ \\ y= \frac{\Delta_y}{\Delta} = \frac{-4}{-4} =1$
$\displaystyle \Delta_z= \left|\begin{array}{ccc}1&1&0\\1&-1&1\\1&1&2\end{array}\right|=\\ \\ =1 \cdot (-1) \cdot 2+0 \cdot 1 \cdot 1+1 \cdot 1 \cdot 1-0 \cdot (-1) \cdot 1-1 \cdot 1 \cdot 2-1 \cdot 1 \cdot 1= \\ \\ =-2+0+1+0-2-1=-4 \\ \\ z= \frac{\Delta_z}{\Delta} = \frac{-4}{-4} =1\\ \\ x=1,~y=1,~z=1$