Răspuns :
[tex]A(m)=\left(\begin{array}{ccc}m & 1 & 1 \\ 2 & m+1 & 1 \\ 1 & 1 & m+1\end{array}\right)[/tex]
a)
Calculam det(A(0)), adaugam primele doua linii ale determinantului si obtinem:
[tex]det(A(0))=\left|\begin{array}{ccc}0 & 1 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|[/tex]
0 1 1
2 1 1
det(A(0))=(0+2+1)-(1+0+2)=3-3=0
b)
m=-3
[tex]\left\{\begin{array}{c}-3 x+y+z=1 \\ 2 x+(-2) y+z=2 \\ x+y+(-2) z=-2\end{array}\right[/tex]
Adunam cele 3 ecuatii si obtinem:
0+0+0=1
0=1 Fals⇒ sistemul de ecuatii nu admite solutii
c)
Calculam det(A(m))
[tex]det(A(m))=\left|\begin{array}{ccc}m & 1 & 1 \\ 2 & m+1 & 1 \\ 1 & 1 & m+1\end{array}\right|[/tex]
m 1 1
2 m+1 1
det(A(m))=[m(m+1)²+2+1]-(m+1+m+2m+2)=m(m+1)²+3-4m-3=m(m+1)²-4m=m[(m+1)²-4]=m(m+1-2)(m+1+2)=m(m-1)(m+3)
Pentru m=0
[tex]\left\{\begin{array}{c}y+z=1 \\ 2 x+ y+z=2 \\ x+y+ z=1\end{array}\right[/tex]
Prima relatie o inlocuim in ultima si obtinem
x+1=1
x=0
Dar daca o inlocuim in a doua obtinem
2x+1=2
2x=1 Ceea ce contrazice ⇒sistemul nu are solutii
Pentru m=1
[tex]\left\{\begin{array}{c} x+y+z=1 \\ 2 x+2 y+z=2 \\ x+y+2z=2\end{array}\right[/tex]
x+y=1-z
Inlocuim in ultima relatie:
1-z+2z=2
1+z=2
z=1
x+y=0
Inlocuim in a doua relatie
2(x+y)+z=2
z=2 Ceea ce contrazice⇒ sistemul nu are solutii
Pentru m=-3 am demonstrat la punctul anterior ca sistemul nu are solutii
Deci pentru m∈R\{-3,0,1} det(A(m))≠0⇒ ca are cel putin o solutie
Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919120
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