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Se consideră matricea [tex]$A(a)=\left(\begin{array}{ccc}1 & -2 a & 0 \\ 0 & 1 & 0 \\ 2 a & -2 a^{2} & 1\end{array}\right)$[/tex], unde [tex]$a$[/tex] este număr real.

[tex]$5 p$[/tex] a) Arătaţi că det [tex]$(A(1))=1$[/tex].

5 p b) Demonstraţi că [tex]$A(a) A(b)=A(a+b)$[/tex], pentru orice numere reale [tex]$a$[/tex] şi [tex]$b$[/tex].

[tex]$5 p$[/tex] c) Demonstrați că, dacă [tex]$A(n)=A(1) A(2) A(3) \cdot \ldots \cdot A(2020)$[/tex], atunci numărul natural [tex]$n$[/tex] este multiplu de 2021 .


Răspuns :

[tex]A(a)=\left(\begin{array}{ccc}1 & -2 a & 0 \\ 0 & 1 & 0 \\ 2 a & -2 a^{2} & 1\end{array}\right)[/tex]

a)

Calculam A(1), inlocuind pe a cu 1

[tex]A(1)=\left(\begin{array}{ccc}1 & -2 & 0 \\ 0 & 1 & 0 \\ 2 & -2 & 1\end{array}\right)[/tex]

Calculam det(A(1)), adaugand primele doua linii

[tex]det(A(1))=\left|\begin{array}{ccc}1 & -2 & 0 \\ 0 & 1 & 0 \\ 2 & -2 & 1\end{array}\right|[/tex]

                      1    -2    0

                      0    1     0

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

b)

[tex]A(a)\times A(b)=A(a+b)[/tex]

[tex]A(a)=\left(\begin{array}{ccc}1 & -2 a & 0 \\ 0 & 1 & 0 \\ 2 a & -2 a^{2} & 1\end{array}\right)\\\\\\A(b)=\left(\begin{array}{ccc}1 & -2 b & 0 \\ 0 & 1 & 0 \\ 2 b & -2 b^{2} & 1\end{array}\right)\\\\\\A(a+b)=\left(\begin{array}{ccc}1 & -2 (a+b) & 0 \\ 0 & 1 & 0 \\ 2 (a+b) & -2 (a+b)^{2} & 1\end{array}\right)[/tex]

Calculam A(a)×A(b)

[tex]A(a)\times A(b)=\left(\begin{array}{ccc}1 & -2 a & 0 \\ 0 & 1 & 0 \\ 2 a & -2 a^{2} & 1\end{array}\right)\times \left(\begin{array}{ccc}1 & -2 b & 0 \\ 0 & 1 & 0 \\ 2 b & -2 b^{2} & 1\end{array}\right)=\left(\begin{array}{ccc}1 & -2 b-2a & 0 \\ 0 & 1 & 0 \\ 2a+2 b & -4ab-2a^2-2b^2 & 1\end{array}\right)[/tex]

-4ab-2a²-2b²=-2(2ab+a²+b²)=-2(a+b)²

[tex]A(a)\times A(b)=\left(\begin{array}{ccc}1 & -2 b-2a & 0 \\ 0 & 1 & 0 \\ 2a+2 b & -4ab-2a^2-2b^2 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & -2(a+b) & 0 \\ 0 & 1 & 0 \\ 2(a+b) & -2(a+b)^2 & 1\end{array}\right)=A(a+b)[/tex]

c)

[tex]A(n)=A(1) A(2) A(3) \cdot \ldots \cdot A(2020)[/tex]

Ne folosim de punctul b, stim ca A(a)A(b)=A(a+b)

Deci A(1)A(2)=A(3)

A(3)A(4)=A(7)

...

[tex]A(1)A(2)A(3)\cdot...\cdot A(2020)=A(1+2+...+2020)=A(\frac{2020\times 2021}{2} )\\\\A(n)=A(\frac{2020\times 2021}{2} )=A(1010\times2021)[/tex]

n=1010×2021, care este multiplu de 2021

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/8858130

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