Question

Se consideră matricea $A(x)=\left(\begin{array}{cc}2^{x} & 0 \\ 0 & 3^{x}\end{array}\right)$, unde $x$ este număr real.

$5 \mathbf{p}$ a) Arătați că $\operatorname{det}(A(x))=6^{x}$, pentru orice număr real $x$

5p b) Determinați numărul real $x$, știind că $A(x) \cdot\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right) \cdot A(x)$.

5 ( c) Demonstraţi că, orice matrice $X \in \mathcal{M}_{2}(\mathbb{R})$ cu proprietatea că $X \cdot X=A(1)$ are două elemente numere iraționale.

Answer 1

$A(x)=\left(\begin{array}{cc}2^{x} & 0 \\ 0 & 3^{x}\end{array}\right)$

a)

Calculam det(A(x)), facem diferenta dintre produsul diagonalelor

det(A(x))=2ˣ×3ˣ-0=6ˣ

b)

$\left(\begin{array}{cc}2^{x} & 0 \\ 0 & 3^{x}\end{array}\right)\cdot \left(\begin{array}{cc}1& 1 \\ 0 &1 \end{array}\right)=\left(\begin{array}{cc}1& 1 \\ 0 &1 \end{array}\right)\cdot \left(\begin{array}{cc}2^{x} & 0 \\ 0 & 3^{x}\end{array}\right)\\\\\left(\begin{array}{cc}2^x&2^x \\ 0 &3^x\end{array}\right)=\left(\begin{array}{cc}2^x& 3^x \\ 0 &3^x\end{array}\right)\\\\2^x=3^x\\\\x=0$

c)

$Fie\ X=\left(\begin{array}{cc}a& b \\ c & d\end{array}\right)\\\\X\cdot X=\left(\begin{array}{cc}a& b \\ c & d\end{array}\right)\cdot \left(\begin{array}{cc}a& b \\ c & d\end{array}\right)=\left(\begin{array}{cc}a^2+bc& b(a+d) \\ c(a+d) & bc+d^2\end{array}\right)$

$\left(\begin{array}{cc}a^2+bc& b(a+d) \\ c(a+d) & bc+d^2\end{array}\right)=\left(\begin{array}{cc}2& 0 \\ 0 &3\end{array}\right)$

a²+bc=2

b(a+d)=0⇒ b=0

c(a+d)=0⇒c=0

bc+d²=3

0+d²=3

d²=3

d=±√3∉R

a²=2

a=±√2∉R

Deci a si d sunt numere irationale

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928531

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