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

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

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

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

Răspuns :

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

a)

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

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

                      2   0   0

                      0   1    2

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

b)

[tex]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)[/tex]

c)

Ne folosim de punctul b

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

[tex]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)[/tex]

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

Dar avem un determinant minor

[tex]\left|\begin{array}{ccc}63&0\\0&12\\\end{array}\right|=756\neq 0[/tex]

Deci rangB=2

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

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