Question

Se consideră sistemul de ecuaţii : mx-2y+z=1
2x-my-3z=3
x-y+2z=4
a) Arătaţi că suma elementelor de pe diagonala principală a matricei sistemului este egală cu 2.
b) Determinaţi valorile reale ale lui m pentru care matricea sistemului are determinantul diferit de zero.
c) Pentru m =1, arătaţi că y1 la a 2= x1 ori z1, unde (x1,y1,z1) este soluţia sistemului.

Answer 1

$\displaystyle \left\{\begin{array}{ccc}mx-2y+z=1\\2x-my-3z=3\\x-y+2z=4\end{array}\right\\ \\ a)A= \left(\begin{array}{ccc}m&-2&1\\2&-m&-3\\1&-1&2\end{array}\right)\\ \\ m+(-m)+2=m-m+2=2$

$\displaystyle b)det(A) \not = 0 \\ \\ det(A)= \left|\begin{array}{ccc}m&-2&1\\2&-m&-3\\1&-1&2\end{array}\right|=\\ \\ =m \cdot (-m) \cdot 2+1 \cdot 2 \cdot (-1)+(-2) \cdot (-3) \cdot 1-1 \cdot (-m) \cdot 1-(-2) \cdot 2\cdot 2-m \cdot (-3) \cdot (-1)=\\ \\ =-2m^2-2+6+m+8-3m=-2m^2-2m+12 \\ \\ -2m^2-2m+12 \not =0 \\ \\ m^2+m-6 \not =0 \\ \\ (m+3)(m-2) \not = 0 \\ \\ m+3 \not =0 \Rightarrow m \not =-3 \\ \\ m-2 \not = 0 \Rightarrow m \not = 2 \\ \\ m \in R-\{-3,2\}$

$\displaystyle c)m=1\\ \\ y_1^2=x_1 \cdot z_1 \\ \\ m=1 \Rightarrow \left\{\begin{array}{ccc}x-2y+z=1\\2x-y-3z=3\\x-y+2z=4\end{array}\right \\ \\ \Delta= \left|\begin{array}{ccc}1&-2&1\\2&-1&-3\\1&-1&2\end{array}\right|=\\ \\ = 1 \cdot (-1) \cdot 2+1 \cdot 2 \cdot (-1)+(-2) \cdot (-3) \cdot 1-1 \cdot (-1) \cdot 1-(-2) \cdot 2 \cdot 2-1 \cdot (-3) \cdot (-1)= \\ \\ =-2-2+6+1+8-3=8$
$\displaystyle \Delta_{x_1}= \left|\begin{array}{ccc}1&-2&1\\3&-1&-3\\4&-1&2\end{array}\right|= \\ \\ =1 \cdot (-1) \cdot 2 +1 \cdot 3 \cdot (-1)+(-2) \cdot (-3) \cdot 4-1 \cdot (-1) \cdot 4-(-2) \cdot 3 \cdot 2-1 \cdot (-3)\cdot (-1)= \\ \\ =-2-3+24+4+12-3=32 \\ \\ x_1= \frac{\Delta_{x_1}}{\Delta} = \frac{32}{8} =4$
$\displaystyle \Delta_{y_1}=\left|\begin{array}{ccc}1&1&1\\2&3&-3\\1&4&2\end{array}\right| =\\ \\ =1 \cdot 3 \cdot 2+1 \cdot 2 \cdot 4+1 \cdot (-3) \cdot 1-1 \cdot 3 \cdot 1-1 \cdot 2 \cdot 2-1 \cdot (-3) \cdot 4= \\ \\ =6+8-3-3-4+12=16 \\ \\ y_1= \frac{\Delta_{y_1}}{\Delta} = \frac{16}{8} =2$
$\displaystyle \Delta_{z_1}= \left|\begin{array}{ccc}1&-2&1\\2&-1&3\\1&-1&4\end{array}\right| = \\ \\ =1 \cdot (-1) \cdot 4+1 \cdot 2 \cdot (-1)+(-2) \cdot 3 \cdot 1-1 \cdot (-1) \cdot 1-(-2) \cdot 2 \cdot 4-1 \cdot 3 \cdot (-1)= \\ \\ =-4-2-6+1+16+3=8 \\ \\ z_1= \frac{\Delta_{z_1}}{\Delta} = \frac{8}{8} =1 \\ \\ x_1=4,~y_1=2,~z_1=1\\ \\ S=\{(4,2,1)\} \\ \\ y_1^2=x_1 \cdot z_1 \Rightarrow 2^2=4 \cdot 1 \Rightarrow 4=4$