Răspuns :
[tex]\displaystyle \mathtt{1.~\left\{\begin{array}{ccc}\mathtt{x+y+mz=m^2-3}\\\mathtt{x-2y+5z=-2}\\\mathtt{3x+2y+(m+1)z=-2}\end{array}\right}\\\\ \mathtt{a)\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt m\\\mathtt1&\mathtt{-2}&\mathtt5\\\mathtt3&\mathtt2&\mathtt{m+1}\end{array}\right|=12}[/tex]
[tex]\displaystyle \mathtt{
\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt
m\\\mathtt1&\mathtt{-2}&\mathtt5\\\mathtt3&\mathtt2&\mathtt{m+1}\end{array}\right|
=1 \cdot (-2) \cdot (m+1)+m\cdot 1\cdot 2+1\cdot 5\cdot 3-m\cdot (-2) \cdot
3-}\\\\\mathtt{-1\cdot1\cdot (m+1)-1\cdot 5\cdot2=-2m-2+2m+15+6m-m-1-10=}\\ \\ \mathtt{=5m+2}\\ \\ \mathtt{5m+2=12\Rightarrow5m=12-2 \Rightarrow 5m=10 \Rightarrow m= \frac{10}{5}\Rightarrow m=2}[/tex]
[tex]\displaystyle \mathtt{b)~x=-3,~y=2,~z=1}\\ \\ \mathtt{ \left \{ {{-3+2+m \cdot
1=m^2-3} \atop {3 \cdot (-3)+2 \cdot 2+(m+1) \cdot 1=-2}} \right. \Rightarrow
\left \{ {{-m^2+m+3-3+2=0} \atop {-9+4+m+1=-2}} \right. \Rightarrow }\\ \\
\mathtt{\Rightarrow \left \{ {{-m^2+m+2=0} \atop {m=2}} \right. }\\ \\
\mathtt{m=2}[/tex]
[tex]\displaystyle \mathtt{c)~m=-1 \Rightarrow
\left\{\begin{array}{ccc}\mathtt{x+y-z=-2}\\\mathtt{x-2y+5z=-2}\\\mathtt{3x+2y=-2}\end{array}\right
\Rightarrow
A=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt{-2}&\mathtt5\\\mathtt3&\mathtt2&\mathtt0\end{array}\right)}[/tex]
[tex]\displaystyle \mathtt{
\Delta=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt{-2}&\mathtt5\\\mathtt3&\mathtt2&\mathtt0\end{array}\right|=1
\cdot (-2) \cdot 0+(-1) \cdot 1 \cdot 2+1 \cdot 5 \cdot 3-}\\ \\ \mathtt{-(-1)
\cdot (-2) \cdot 3-1 \cdot 1 \cdot 0-1 \cdot 5 \cdot 2=-2+15-6-10=-3}\\ \\
\mathtt{\Delta=-3 \not =0}[/tex]
[tex]\displaystyle \mathtt{\Delta_x=
\left|\begin{array}{ccc}\mathtt{-2}&\mathtt1&\mathtt{-1}\\\mathtt{-2}&\mathtt{-2}&\mathtt5\\\mathtt{-2}&\mathtt2&\mathtt0\end{array}\right|=(-2)
\cdot (-2) \cdot 0+(-1) \cdot (-2)\cdot 2+1 \cdot 5 \cdot (-2) -}\\ \\
\mathtt{-(-1) \cdot (-2) \cdot (-2)-1 \cdot (-2) \cdot 0-(-2) \cdot 5 \cdot
2=4-10+4+20=18}\\ \\ \mathtt{\Delta_x=18}[/tex]
[tex]\displaystyle \mathtt{\Delta_y=
\left|\begin{array}{ccc}\mathtt1&\mathtt{-2}&\mathtt{-1}\\\mathtt1&\mathtt{-2}&\mathtt5\\\mathtt3&\mathtt{-2}&\mathtt0\end{array}\right|=1
\cdot (-2) \cdot 0+(-1) \cdot 1 \cdot (-2)+(-2) \cdot 5 \cdot 3-}\\ \\
\mathtt{-(-1) \cdot (-2) \cdot 3-(-2) \cdot 1 \cdot 0-1\cdot5\cdot(-2)=2-30-6+10=-24}\\ \\ \mathtt{\Delta_y=-24}[/tex]
[tex]\displaystyle \mathtt{\Delta_z=\left|\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt{-2}\\\mathtt1&\mathtt{-2}&\mathtt{-2}\\\mathtt3&\mathtt2&\mathtt{-2}\end{array}\right|=1\cdot(-2)\cdot(-2)+(-2)\cdot1\cdot 2+1 \cdot (-2) \cdot 3-}\\ \\
\mathtt{-(-2)\cdot(-2)\cdot3-1\cdot1\cdot (-2)-1 \cdot (-2) \cdot
2=4-4-6-12+2+4=-12}\\\\\mathtt{\Delta_z=-12}[/tex]
[tex]\displaystyle \mathtt{x= \frac{\Delta_x}{\Delta}=\frac{18}{-3}=-6}\\\\ \mathtt{y= \frac{\Delta_y}{\Delta}= \frac{-24}{-3}=8}\\\\\mathtt{z= \frac{\Delta_z}{\Delta}= \frac{-12}{-3}=4}\\ \\
\mathtt{x=-6;~y=8;~z=4}[/tex]
[tex]\displaystyle \mathtt{2.~\left\{\begin{array}{ccc}\mathtt{x+ay+a^2z=a+1}\\\mathtt{x+by+b^2z=b+1}\\\mathtt{x+cy+c^2z=c+1}\end{array}\right
}\\ \\ \mathtt{a)~a=-1;~b=0;~c=2\Rightarrow \left
\left\{\begin{array}{ccc}\mathtt{x-y+z=0}\\\mathtt{x=1}\\\mathtt{x+2y+4z=3}\end{array}\right.\Rightarrow\left \{ {{x-y+z=0} \atop {x+2y+4z=3}} \right. \Rightarrow}[/tex]
[tex]\displaystyle\mathtt{\Rightarrow \left \{{{1-y+z=0}\atop{1+2y+4z=3}} \right. \Rightarrow\left\{{{-y+z=-1|\cdot 2}\atop
{2y+4z=2}}\right.\Rightarrow \left \{ {{-2y+2z=-2}\atop{2y+4z=2}}\right.}\\
\mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~--------}\\
\mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~/~~~~6z=0\Rightarrow z=0}\\\\\mathtt{2y+4z=2 \Rightarrow 2y+4 \cdot0=2\Rightarrow2y=2\Rightarrow y=1}\\\\ \mathtt{x=1;~y=1;~z=0}[/tex]
[tex]\displaystyle \mathtt{b)~det(A)=(a-b)(b-c)(c-a)}\\ \\
\mathtt{\left\{\begin{array}{ccc}\mathtt{x+ay+a^2z=a+1}\\\mathtt{x+by+b^2z=b+1}\\\mathtt{x+cy+c^2z=c+1}\end{array}\right\Rightarrow A=\left(\begin{array}{ccc}\mathtt 1&\mathtt a&\mathtt
a^2\\\mathtt1&\mathtt b&\mathtt b^2\\\mathtt1&\mathtt
c&\mathtt c^2\end{array}\right)}[/tex]
[tex]\displaystyle \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt 1&\mathtt a&\mathtt a^2\\\mathtt1&\mathtt b&\mathtt b^2\\\mathtt 1&\mathtt c&\mathtt c^2\end{array}\right|=\left|\begin{array}{ccc}\mathtt 1&\mathtt a&\mathtt a^2\\\mathtt0&\mathtt{b-a}&\mathtt{b^2-a^2}\\\mathtt 0&\mathtt {c-a}&\mathtt{c^2-a^2}\end{array}\right| }\\ \\ \mathtt{b^2-a^2=(b-a)(b+a);~c^2-a^2=(c-a)(c+a)}[/tex]
[tex]\displaystyle \mathtt{\left|\begin{array}{ccc}\mathtt 1&\mathtt a&\mathtt a^2\\\mathtt0&\mathtt{b-a}&\mathtt{b^2-a^2}\\\mathtt 0&\mathtt {c-a}&\mathtt{c^2-a^2}\end{array}\right|=(b-a)(c-a) \left|\begin{array}{ccc}\mathtt1&\mathtt{b+a}\\\mathtt1&\mathtt{c+a}\end{array}\right|=}\\ \\ \mathtt{=(b-a)(c-a)(c+a-b-a)=(b-a)(c-a)(c-b)=}\\ \\ \mathtt{=(a-b)(b-c)(c-a)}[/tex]
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