Răspuns :
[tex]A(a)=\left(\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & a \\ 1 & -a & 0\end{array}\right)[/tex]
[tex](A(a))^{t}=\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -a \\ 3 & a & 0\end{array}\right)[/tex]
a)
Calculam det(A(2)), inlocuind pe a cu 2 si adaugand primele doua linii ale determinantului
[tex]det(A(2))=\left|\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & 2 \\ 1 & -2 & 0\end{array}\right|[/tex]
1 0 3
0 1 2
det(A(2))=(0+0+0)-(3-4+0)=0+1=1
b)
Ca o matrice sa fie inversabila trebuie ca determinantul sau sa fie diferit de 0
[tex]det(A(q))=\left|\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & q \\ 1 & -q & 0\end{array}\right|[/tex]
1 0 3
0 1 q
det(A(q))=(0+0+0)-(3-q²+0)=q²-3
Daca q²-3=0, atunci q=±√3, ceea ce nu verifica, deoarece q este numar rational⇒ q²-3≠0⇒ matricea este inversabila
c)
Calculam:
[tex]A(a)-(A(a))^t=\left(\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & a \\ 1 & -a & 0\end{array}\right)-\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -a \\ 3 & a & 0\end{array}\right)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2a \\ -2 & -2a & 0\end{array}\right)[/tex]
[tex]B(a)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2a \\ -2 & -2a & 0\end{array}\right)\\\\\\B(p)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)[/tex]
[tex]B(p)B(p)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)\times \left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)=\left(\begin{array}{ccc}-4 & -4p & 0\\ -4p & -4p^2 & 0 \\ 0 & 0 & -4-4p^2\end{array}\right)[/tex]
[tex]B(p)B(p)B(p)=\left(\begin{array}{ccc}-4 & -4p & 0\\ -4p & -4p^2 & 0 \\ 0 & 0 & -4-4p^2\end{array}\right)\times \left(\begin{array}{ccc}0 & 0 & 2\\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)=\\\\\\=\left(\begin{array}{ccc}0 & 0 & -8-8p^2\\0 & 0 & -8p-8p^3 \\ 8+8p^2 &8p+8p^3 & 0\end{array}\right)[/tex]
[tex]\left(\begin{array}{ccc}0 & 0 & -8-8p^2\\0 & 0 & -8p-8p^3 \\ 8+8p^2 &8p+8p^3 & 0\end{array}\right)+5\left(\begin{array}{ccc}0 & 0 & 2\\0 & 0 & 2p \\ -2 &-2p & 0\end{array}\right)=\\\\\\\\\\=\left(\begin{array}{ccc}0 & 0 & 2-8p^2\\0 & 0 & 2p-8p^3 \\ -2+8p^2 &-2p+8p^3 & 0\end{array}\right)=O_3[/tex]
2-8p²=0
8p²=2
[tex]p^2=\frac{1}{4}\\\\ p=\frac{1}{2}\ \ si \\\ p=-\frac{1}{2}[/tex]
Un exercitiu similar cu matrice gasesti aici: https://brainly.ro/tema/4677939
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