Question

Se consideră matricele $A(x)=\left(\begin{array}{cc}x & 3 \\ -3 & x\end{array}\right)$, unde $x$ este număr real.

$5 \mathbf{p}$ a) Arătați că $\operatorname{det}(A(x))=x^{2}+9$, pentru orice număr real $x$.

5p b) Demonstrați că $A(2020-x)+A(2020+x)=2 A(2020)$, pentru orice număr real x .

$5 p$ c) Determinați numărul natural $n$, pentru care $A(n) A(2-n)=2 A(-6)$.

Answer 1

$A(x)=\left(\begin{array}{cc}x & 3 \\ -3 & x\end{array}\right)$

a)

Calculam det(A(x)), luam produsul primei diagonale si scadem produsul celei de-a doua diagonale

$det(A(x))=\left|\begin{array}{cc}x & 3 \\ -3 & x\end{array}\right|=x^2-[3\times(-3)]=x^2+9$

b)

Calculam A(2020-x), A(2020+x) si A(2020)

$A(2020-x)=\left(\begin{array}{cc}2020-x & 3 \\ -3 & 2020-x\end{array}\right)$

$A(2020+x)=\left(\begin{array}{cc}2020+x & 3 \\ -3 & 2020+x\end{array}\right)$

$A(2020)=\left(\begin{array}{cc}2020 & 3 \\ -3 & 2020\end{array}\right)$

Vom calcula A(2020-x)+A(2020+x)

$A(2020-x)+A(2020+x)=\left(\begin{array}{cc}2020-x & 3 \\ -3 & 2020-x\end{array}\right)+\left(\begin{array}{cc}2020+x & 3 \\ -3 & 2020+x\end{array}\right)$

Se aduna termen cu termen si obtinem:

$A(2020-x)+A(2020+x)=\left(\begin{array}{ccc}4040&6\\-6&4040\end{array}\right)$

Dam factor comun pe 2 si obtinem:

$A(2020-x)+A(2020+x)=2\left(\begin{array}{ccc}2020&3\\-3&2020\end{array}\right)=2A(2020)$

c)

Calculam A(n)×A(2-n) si apoi egalam cu 2A(-6)

$A(n)\times A(2-n)=\left(\begin{array}{cc}n & 3 \\ -3 & n\end{array}\right)\times \left(\begin{array}{cc}2-n & 3 \\ -3 &2- n\end{array}\right)\\\\A(n)\times A(2-n)=\left(\begin{array}{cc}2n-n^2-9 & 3n+6-3n \\ -6+3n-3n & -9+2n-n^2\end{array}\right)$

$\left(\begin{array}{cc}2n-n^2-9 & 3n+6-3n \\ -6+3n-3n & -9+2n-n^2\end{array}\right)=2\left(\begin{array}{cc}-6 & 3 \\ -3 & -6\end{array}\right)$

Introducem 2 in matrice, inmultind fiecare termen:

$\left(\begin{array}{cc}2n-n^2-9 & 3n+6-3n \\ -6+3n-3n & -9+2n-n^2\end{array}\right)=\left(\begin{array}{cc}-12 & 6 \\ -6 & -12\end{array}\right)$

$\left(\begin{array}{cc}2n-n^2-9 & 6 \\ -6& -9+2n-n^2\end{array}\right)=\left(\begin{array}{cc}-12 & 6 \\ -6 & -12\end{array}\right)$

Se egaleaza termenii si obtinem:

-n²+2n-9=-12 |×(-1)

n²-2n+9=12

n²-2n-3=0

n²-2n-2-1=0

n²-1-2(n+1)=0

(n-1)(n+1)-2(n+1)=0

(n+1)(n-3)=0

n=-1 si n=3

n=-1∉N

Deci solutia finala este n=3

Un exercitiu similar de bac gasesti aici:https://brainly.ro/tema/7350274

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