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Se consideră matricele [tex]$A=\left(\begin{array}{cc}1 & 3 \\ 2 & 1\end{array}\right), B=\left(\begin{array}{cc}-4 & 0 \\ 0 & 4\end{array}\right)$[/tex] şi [tex]$M(x)=\left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)$[/tex], unde [tex]$x$[/tex] este număr real.

[tex]$5 p$[/tex] a) Arătați că det [tex]$A=-5$[/tex].

[tex]$5 \mathbf{p}$[/tex] b) Arătați că [tex]$\operatorname{det}(A+M(-1))=\operatorname{det} B$[/tex].

[tex]$5 p$[/tex] c) Determinați numărul real [tex]$x$[/tex] pentru care [tex]$M(x) \cdot A-A \cdot M(x)=B$[/tex].

Răspuns :

[tex]A=\left(\begin{array}{cc}1 & 3 \\ 2 & 1\end{array}\right)[/tex]

[tex]B=\left(\begin{array}{cc}-4 & 0 \\ 0 & 4\end{array}\right)[/tex]

[tex]M(x)=\left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)[/tex]

a)

Facem diferenta dintre produsul diagonalelor si obtinem:

detA=1-6=-5

b)

det(A+M(-1))=detB

detB=-16-0=-16

[tex]det(A+M(-1))=\left|\begin{array}{ll}0 & 4 \\ 4 &4\end{array}\right|=0-16=-16[/tex]

Observam ca sunt egale

c)

[tex]M(x)\cdot A=\left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)\cdot \left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)=\left(\begin{array}{ll}x+2 & 3x+1 \\ 8 & 9\end{array}\right)[/tex]

[tex]A\cdot M(x)= \left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)\cdot \left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)=\left(\begin{array}{ll}x+6 & 10\\ 2x+2 & 5\end{array}\right)[/tex]

[tex]\left(\begin{array}{ll}x+2 & 3x+1\\ 8 & 9\end{array}\right)-\left(\begin{array}{ll}x+6 & 10\\ 2x+2 & 5\end{array}\right)=\left(\begin{array}{ll}-4 & 3x-9\\ -2x+6 & 4\end{array}\right)[/tex]

Egalam cu termenii matricei B si obtinem

3x-9=0

3x=9

x=3

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928396

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