a)
x=0
y=3
z=1
Inlocuim si obtinem:
0-6+3=-6+3=-3 (A)
0+3+1=4 (A)
0-3+4=-3+4=1 (A)
b)
Pentru ca sistemul sa admita solutie unica, atunci determinantul trebuie sa fie diferit de 0
[tex]\Delta=\left|\begin{array}{ccc}1&-2&3\\2&1&1\\m&-1&4\end{array}\right|[/tex]
1 -2 3
2 1 1
Δ=(4-6-2m)-(3m-1-16)=-5m+15
-5m+15≠0
5m≠16
m≠3
m∈R\{3}
c)
m≠3
Metoda lui Cramer
[tex]\Delta_x=\left|\begin{array}{ccc}-3&-2&3\\4&1&1\\1&-1&4\end{array}\right|[/tex]
-3 -2 3
4 1 1
[tex]\Delta_x=-12-12-2-(3+3-32)=-26-6+32=0\\\\x=\frac{\Delta_x}{\Delta}=0[/tex]
Am inlocuit coloana coeficientilor lui x cu coloana termenilor liberi
[tex]\Delta_y=\left|\begin{array}{ccc}1&-3&3\\2&4&1\\m&1&4\end{array}\right|[/tex]
1 -3 3
2 4 1
[tex]\Delta_y=16+6-3m-(12m+1-24)=22-3m-12m+23=-15m+45=3(-5m+15)\\\\y=3[/tex]
Am inlocuit coloana coeficientilor lui y cu coloana termenilor liberi
[tex]\Delta_z=\left|\begin{array}{ccc}1&-2&-3\\2&1&4\\m&-1&1\end{array}\right|[/tex]
1 -2 -3
2 1 4
[tex]\Delta_z=1+6-8m-(-3m-4-4)=-5m+15\\\\z=1[/tex]
Am inlocuit coloana coeficientilor lui z cu coloana termenilor liberi
Un alt exercitiu gasesti aici: https://brainly.ro/tema/9685683
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