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Se consideră matricele [tex]$A=\left(\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right)$[/tex] şi [tex]$B(x, y)=\left(\begin{array}{cc}x & 1 \\ y & -1\end{array}\right)$[/tex], unde [tex]$x$[/tex] şi [tex]$y$[/tex] sunt numere reale.

5p a) Arătați că [tex]$\operatorname{det} A=-4$[/tex].

[tex]$5 p$[/tex] b) Arătați că [tex]$\operatorname{det}(A-2 B(x, y))=0$[/tex], pentru orice numere reale [tex]$x$[/tex] şi [tex]$y$[/tex].

[tex]$5 p$[/tex] c) Determinați numerele reale [tex]$x$[/tex] şi [tex]$y$[/tex] pentru care [tex]$A \cdot B(x, y)=B(x, y) \cdot A$[/tex].


Răspuns :

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

[tex]B(x, y)=\left(\begin{array}{cc}x & 1 \\ y & -1\end{array}\right)[/tex]

a)

det(A)=-4

Calculam detA, facand diferenta dintre produsul diagonalelor determinantului:

[tex]det(A)=\left|\begin{array}{ccc}1&2\\1&-2\\\end{array}\right|=-2-2=-4[/tex]

b)

[tex]{det}(A-2 B(x, y))=0[/tex]

[tex]A-2 B(x, y)=\left(\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right)-\left(\begin{array}{cc}2x & 2 \\ 2y & -2\end{array}\right)=B(x, y)=\left(\begin{array}{cc}1-2x & 0 \\ 1-2y & 0\end{array}\right)[/tex]

[tex]det=\left|\begin{array}{cc}1-2x & 0 \\ 1-2y & 0\end{array}\right|=0-0=0[/tex]

c)

[tex]A\cdot B(x,y)=\left(\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right)\cdot \left(\begin{array}{cc}x & 1 \\ y& -1\end{array}\right)=\left(\begin{array}{cc}x +2y& -1 \\ x-2y&3\end{array}\right)[/tex]

[tex]B(x,y)\cdot A= \left(\begin{array}{cc}x & 1 \\ y& -1\end{array}\right)\cdot \left(\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right)=\left(\begin{array}{cc}x+1 & 2x-2 \\ y-1 &2y+2\end{array}\right)[/tex]

Egalam termenii si obtinem:

x+2y=x+1

2y=1

[tex]y=\frac{1}{2}[/tex]

-1=2x-2

2x=1

[tex]x=\frac{1}{2}[/tex]

Un alt exercitiu similar de bac il gasesti aici:https://brainly.ro/tema/2494494

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