Question

Se consideră matricele $A(x)=\left(\begin{array}{ccc}2^{x} & 0 & 0 \\ 0 & 1 & 2 x \\ 0 & 0 & 1\end{array}\right)$ şi $I_{3}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$, unde $x$ este număr real.

$5 p$ a) Arătați că $\operatorname{det}(A(1))=2$.

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

$5 p$ c) Determinați rangul matricei $B=A(1) \cdot A(2) \cdot A(3)-I_{3}$.

Answer 1

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

a)

Calculam det(A(1)), adaugam primele doua linii ale determinantului si obtinem:

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

                      2   0   0

                      0   1    2

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

b)

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

c)

Ne folosim de punctul b

A(1)A(2)A(3)=A(1+2+3)=A(6)

$B=\left(\begin{array}{ccc}2^{6} & 0 & 0 \\ 0 & 1 & 12 \\ 0 & 0 & 1\end{array}\right)-\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}63& 0 & 0 \\ 0 & 0& 12 \\ 0 & 0 & 0\end{array}\right)$

detB=0 (ultima linie este 0 0 0 ) ⇒rangB≤2

Dar avem un determinant minor

$\left|\begin{array}{ccc}63&0\\0&12\\\end{array}\right|=756\neq 0$

Deci rangB=2

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919124

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