Question

Se consideră matricea $A(x)=\left(\begin{array}{cc}1+x & x \\ 2 x & 1+2 x\end{array}\right)$, unde $x$ este număr real.

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

5p b) Demonstrați că $A(x) A(y)=A(x+y+3 x y)$, pentru orice numere reale $x$ şi $y$.

$5 p$ c) Determinați numerele reale a pentru care $A(a) A(a)=A(5)$.

Answer 1

$A(x)=\left(\begin{array}{cc}1+x & x \\ 2 x & 1+2 x\end{array}\right)$

a)

Calculam det(A(1)), inlocuind pe a cu 1  si facem diferenta dintre produsul diagonalelor:

$det(A(1))=\left|\begin{array}{cc}2& 1 \\ 2 & 3\end{array}\right|=6-2=4$

b)

$A(x)A(y)=\left(\begin{array}{cc}1+x & x \\ 2 x & 1+2 x\end{array}\right)\times \left(\begin{array}{cc}1+y & y \\ 2 y & 1+2 y\end{array}\right)=\\\\\\=\left(\begin{array}{cc}(1+x)(1+y)+2xy& y+xy+x+2xy \\ 2x+2xy+2y+4xy & 2xy+(1+2x)(1+2y)\end{array}\right)=\\\\=\left(\begin{array}{cc}1+y+x+xy+2xy&x+y+3xy \\ 2x+2y+6xy & 2xy+1+2y+2x+4xy\end{array}\right)=\\\\\left(\begin{array}{cc}1+x+y+3xy&x+y+3xy \\ 2(x+y+3xy) & 1+2(x+y+3xy\end{array}\right)=A(x+y+3xy)$

c)

Ne folosim de punctul b

A(x)A(y)=A(x+y+3xy)

A(a)A(a)=A(a+a+3a×a)

A(2a+3a²)=A(5)

3a²+2a=5

3a²+2a-5=0

Δ=4+60=64

$a_1=\frac{-2+8}{6} =1\\\\a_2=\frac{-2-8}{6} =-\frac{5}{3}$

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/6937428

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