Question

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

$5 \mathbf{p}$ a) Arătați că $\operatorname{det}(A(1))=-4$.

5p b) Demonstrați că $\operatorname{det}(A(x) A(y)-A(2 x y))=0$, pentru orice numere reale $x$ și $y$.

$5 p$ c) Determinați numărul natural $n$ pentru care $A(1) A\left(\frac{1}{2}\right)+A(2) A\left(\frac{1}{4}\right)+\ldots+A(1010) A\left(\frac{1}{2020}\right)=n I_{3}$.

Answer 1

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

a)

Calculam A(1), inlocuind pe x cu 1, apoi calculam det(A(1)), adaugand primele doua linii

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

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

                       1      1    -1

                       1      0    1

det(A(1))=(0-1-1)-(0+1+1)=-2-2=-4

b)

Demonstrati ca ${det}(A(x) A(y)-A(2 x y))=0$

Calculam prima data A(x)A(y)

$A(x)A(y)=\left(\begin{array}{ccc}x & 1 & -x \\ 1 & 0 & 1 \\ -x & 1 & x\end{array}\right)\times \left(\begin{array}{ccc}y & 1 & -y \\ 1 & 0 & 1 \\ -y & 1 & y\end{array}\right)=\\\\\left(\begin{array}{ccc}2xy+1 &0 & -2xy+1 \\ 0 & 2 & 0 \\ -2xy+1 & 0 & 2xy+1\end{array}\right)$

Apoi calculam A(x)A(y)-A(2xy)

$A(x)A(y)-A(2xy)=\left(\begin{array}{ccc}2xy+1 &0 & -2xy+1 \\ 0 & 2 & 0 \\ -2xy+1 & 0 & 2xy+1\end{array}\right)-\left(\begin{array}{ccc}2xy &1 & -2xy \\ 1 & 0 & 1 \\ -2xy & 1 & 2xy\end{array}\right)=\left(\begin{array}{ccc}1 &-1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 1\end{array}\right)$

Calculam det(A(x)A(y)-A(2xy))

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

                                              1      -1       1

                                             -1      2      -1

det(A(x)A(y)-A(2xy))=(2+1+1)-(2+1+1)=4-4=0

c)

$A(1) A\left(\frac{1}{2}\right)+A(2) A\left(\frac{1}{4}\right)+\ldots+A(1010) A\left(\frac{1}{2020}\right)=n I_{3}$

Ne folosim de punctul b, stim ca

$A(x)A(y)=\left(\begin{array}{ccc}2xy+1 &0 & -2xy+1 \\ 0 & 2 & 0 \\ -2xy+1 & 0 & 2xy+1\end{array}\right)$

Observam ca daca $y=\frac{1}{2x}$ atunci

$A(x)A(y)=A(x)A(\frac{1}{2x})= \left(\begin{array}{ccc}2 &0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right)$

Dam factor comun pe 2 si vom obtine:

$A(x)A(y)=A(x)A(\frac{1}{2x})= 2\times \left(\begin{array}{ccc}1 &0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)=2I_3$

$A(1) A\left(\frac{1}{2}\right)+A(2) A\left(\frac{1}{4}\right)+\ldots+A(1010) A\left(\frac{1}{2020}\right)=n I_{3}$

Observam ca fiecare termen va fi egal cu 2I₃. Avem 1010 termeni deci suma noastra va fi egala cu:

$A(1) A\left(\frac{1}{2}\right)+A(2) A\left(\frac{1}{4}\right)+\ldots+A(1010) A\left(\frac{1}{2020}\right)=1010\times 2I_3$

$1010\times 2I_3=nI_3$

De aici rezulta ca n=2020

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/1009519

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