Question

1.Fie sistemul e ecuații
(mx+y+z=m la puterea 2 - 3
(5x-2y+z=-2, m e R
((m+1)x + 2y+3z=-2
A). Determinați m e R ,știind că det(A)=-12;
B). Pentru m e R, așa încât sistemul să aibă soluția (1 ,2 , -3);
C). Pentru m = -1, rezolvați sistemul de ecuații.

Answer 1

$\displaystyle \mathtt{ \left\{\begin{array}{ccc}\mathtt{mx+y+z=m^2-3}\\\mathtt{5x-2y+z=-2}\\\mathtt{(m+1)x+2y+3z=-2}\end{array}\right,~m\in\mathbb{R}~~~~~~~~~~~~A= \left(\begin{array}{ccc}\mathtt m&\mathtt1&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\mathtt{m+1}&\mathtt2&\mathtt3\end{array}\right)}$

$\displaystyle \mathtt{a)~det(A)=-12}\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt m&\mathtt1&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\mathtt{m+1}&\mathtt2&\mathtt3\end{array}\right|=m\cdot(-2)\cdot3+1\cdot5\cdot2+1\cdot1\cdot(m+1)-}\\ \\ \mathtt{-~1\cdot(-2)\cdot(m+1)-1\cdot5\cdot3-m\cdot1\cdot2=-6m+10+m+1+2m+2-}\\ \\ \mathtt{-15-2m=-5m-2}\\ \\ \mathtt{-5m-2=-12\Rightarrow-5m=-12+2\Rightarrow-5m=-10\Rightarrow m=2}$

$\displaystyle\mathtt{b)~x=1;~y=2;~z=-3}\\\\\mathtt{\left\{{{mx+y+z=m^2-3} \atop{(m+1)x+2y+3z=-2}}\right.\Rightarrow\left\{{{m\cdot1+2-3=m^2-3}\atop {(m+1)\cdot1+2\cdot2+3\cdot(-3)=-2}}\right.\Rightarrow }\\\\ \mathtt{\Rightarrow\left\{{{m-1=m^2-3}\atop{m+1+4-9=-2}}\right. \Rightarrow\left\{{{-m^2+m-1+3=0}\atop{m-4=-2}}\right.\Rightarrow}\\\\ \mathtt{\Rightarrow\left\{{{-m^2+m+2=0}\atop{m=-2+4}}\right.\Rightarrow\left\{{{-m^2+m+2=0}\atop{m=2}}\right.}\\\\\mathtt{m=2}$

$\displaystyle \mathtt{c)~m=-1 \Rightarrow \left\{\begin{array}{ccc}\mathtt{-x+y+z=-2}\\\mathtt{5x-2y+z=-2}\\\mathtt{2y+3z=-2}\end{array}\right}\\ \\ \mathtt{\Delta=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt2&\mathtt3\end{array}\right|=(-1)\cdot(-2)\cdot3+1\cdot5\cdot2+1\cdot1\cdot0-1\cdot(-2)\cdot0-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1\cdot5\cdot3-(-1)\cdot1\cdot2=3}\\ \\ \mathtt{\Delta=3}$

$\displaystyle \mathtt{\Delta_x= \left|\begin{array}{ccc}\mathtt{-2}&\mathtt1&\mathtt1\\\mathtt{-2}&\mathtt{-2}&\mathtt1\\\mathtt{-2}&\mathtt2&\mathtt3\end{array}\right|=(-2)\cdot(-2)\cdot3+1\cdot(-2)\cdot2+1\cdot1\cdot(-2)-}\\ \\ \mathtt{-1\cdot(-2)\cdot(-2)-1\cdot(-2)\cdot3-(-2)\cdot1\cdot2=12}\\ \\ \mathtt{\Delta_x=12}$

$\displaystyle \mathtt{\Delta_y= \left|\begin{array}{ccc}\mathtt{-1}&\mathtt{-2}&\mathtt1\\\mathtt5&\mathtt{-2}&\mathtt1\\\mathtt0&\mathtt{-2}&\mathtt3\end{array}\right|=(-1)\cdot(-2)\cdot3+1\cdot5\cdot(-2)+(-2)\cdot1\cdot0-}\\ \\ \mathtt{-1\cdot(-2)\cdot0-(-2)\cdot5\cdot3-(-1)\cdot1\cdot(-2)=24}\\ \\ \mathtt{\Delta_y=24}$

$\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt{-1}&\mathtt1&\mathtt{-2}\\\mathtt5&\mathtt{-2}&\mathtt{-2}\\\mathtt0&\mathtt2&\mathtt{-2}\end{array}\right|=(-1)\cdot(-2)\cdot(-2)+(-2)\cdot5\cdot2+1\cdot(-2)\cdot0-}\\ \\ \mathtt{-(-2)\cdot(-2)\cdot0-1\cdot5\cdot(-2)-(-1)\cdot(-2)\cdot2=-18}\\ \\ \mathtt{\Delta_z=-18}$

$\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}= \frac{12}{3}=4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=4}\\ \\ \mathtt{y= \frac{\Delta_y}{\Delta} = \frac{24}{3}=8~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y=8}\\ \\ \mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-18}{3}=-6~~~~~~~~~~~~~~~~~~~~~~~~~z=-6}$