Question

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

$5 p$ a) Arătaţi că det $(A(1))=1$.

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

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

Answer 1

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

a)

Calculam A(1), inlocuind pe a cu 1

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

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

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

                      1    -2    0

                      0    1     0

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

b)

$A(a)\times A(b)=A(a+b)$

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

Calculam A(a)×A(b)

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

-4ab-2a²-2b²=-2(2ab+a²+b²)=-2(a+b)²

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

c)

$A(n)=A(1) A(2) A(3) \cdot \ldots \cdot A(2020)$

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)

...

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

n=1010×2021, care este multiplu de 2021

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

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