Question

Se consideră matricele $O_{3}=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right), A(a)=\left(\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & a \\ 1 & -a & 0\end{array}\right)$ şi $(A(a))^{t}=\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -a \\ 3 & a & 0\end{array}\right)$, unde $a$ este număr real.

$5 \mathbf{p}$ a) Arătaţi că det $(A(2))=1$.

$5 p$ b) Demonstrați că, pentru orice număr rațional $q$, matricea $A(q)$ este inversabilă.

$5 p$ c) Se consideră matricea $B(a)=A(a)-(A(a))^{t}$. Determinați numerele raționale $p$ pentru care $B(p) B(p) B(p)+5 B(p)=O_{3}$

Answer 1

$A(a)=\left(\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & a \\ 1 & -a & 0\end{array}\right)$

$(A(a))^{t}=\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -a \\ 3 & a & 0\end{array}\right)$

a)

Calculam det(A(2)), inlocuind pe a cu 2 si adaugand primele doua linii ale determinantului

$det(A(2))=\left|\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & 2 \\ 1 & -2 & 0\end{array}\right|$

                      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

$det(A(q))=\left|\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & q \\ 1 & -q & 0\end{array}\right|$

                      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:

$A(a)-(A(a))^t=\left(\begin{array}{ccc}1 & 0 & 3 \\ 0 & 1 & a \\ 1 & -a & 0\end{array}\right)-\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -a \\ 3 & a & 0\end{array}\right)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2a \\ -2 & -2a & 0\end{array}\right)$

$B(a)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2a \\ -2 & -2a & 0\end{array}\right)\\\\\\B(p)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)$

$B(p)B(p)=\left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)\times \left(\begin{array}{ccc}0 & 0 & 2 \\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)=\left(\begin{array}{ccc}-4 & -4p & 0\\ -4p & -4p^2 & 0 \\ 0 & 0 & -4-4p^2\end{array}\right)$

$B(p)B(p)B(p)=\left(\begin{array}{ccc}-4 & -4p & 0\\ -4p & -4p^2 & 0 \\ 0 & 0 & -4-4p^2\end{array}\right)\times \left(\begin{array}{ccc}0 & 0 & 2\\ 0 & 0 & 2p \\ -2 & -2p & 0\end{array}\right)=\\\\\\=\left(\begin{array}{ccc}0 & 0 & -8-8p^2\\0 & 0 & -8p-8p^3 \\ 8+8p^2 &8p+8p^3 & 0\end{array}\right)$

$\left(\begin{array}{ccc}0 & 0 & -8-8p^2\\0 & 0 & -8p-8p^3 \\ 8+8p^2 &8p+8p^3 & 0\end{array}\right)+5\left(\begin{array}{ccc}0 & 0 & 2\\0 & 0 & 2p \\ -2 &-2p & 0\end{array}\right)=\\\\\\\\\\=\left(\begin{array}{ccc}0 & 0 & 2-8p^2\\0 & 0 & 2p-8p^3 \\ -2+8p^2 &-2p+8p^3 & 0\end{array}\right)=O_3$

2-8p²=0

8p²=2

$p^2=\frac{1}{4}\\\\ p=\frac{1}{2}\ \ si \\\ p=-\frac{1}{2}$

Un exercitiu similar cu matrice gasesti aici: https://brainly.ro/tema/4677939

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