Răspuns :
[tex]A(m, x)=\left(\begin{array}{ccc}2 & -x & 1 \\ 1 & m & 3 \\ 3 & -2 & x\end{array}\right)[/tex]
a)
det(A(4,2))=0
Calculam det(A(4,2)), inlocuim pe m cu 4 si pe x cu 2, punem primele doua linii ale determinantului si obtinem:
[tex]det(A(4, 2))=\left|\begin{array}{ccc}2 & -2 & 1 \\ 1 & 4 & 3 \\ 3 & -2 & 2\end{array}\right|[/tex]
2 -2 1
1 4 3
det(A(4,2))=(16-2-18)-(12-12-4)=-4+4=0
b)
[tex]A(2, 1)=\left(\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & -2 & 1\end{array}\right)[/tex]
Calculam det(A(2,1))
[tex]det(A(2, 1))=\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & -2 & 1\end{array}\right|[/tex]
2 -1 1
1 2 3
det(A(2,1))=(4-2-9)-(6-12-1)=-7+7=0⇒ rang≤2
[tex]\left|\begin{array}{ccc}2&-1\\1&2\\\end{array}\right|=4+1=5\neq 0[/tex] ⇒ rang=2
c)
det(A(3,n))=det(A(3,p))
[tex]det(A(3, n))=\left|\begin{array}{ccc}2 & -n & 1 \\ 1 & 3 & 3 \\ 3 & -2 & n\end{array}\right|[/tex]
2 -n 1
1 3 3
det(A(3,n))=(6n-2-9n)-(9-12-n²)=n²-3n+1
Deci n²-3n+1=p²-3p+1
(n-p)(n+p-3)=0
n=p nu se poate
n+p=3
n=1 si p=2 sau n=2 si p=1
Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9905463
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