Question

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

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

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

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

Answer 1

$A=\left(\begin{array}{cc}1 & 2 \\ 1 & -2\end{array}\right)$

$B(x, y)=\left(\begin{array}{cc}x & 1 \\ y & -1\end{array}\right)$

a)

det(A)=-4

Calculam detA, facand diferenta dintre produsul diagonalelor determinantului:

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

b)

${det}(A-2 B(x, y))=0$

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

$det=\left|\begin{array}{cc}1-2x & 0 \\ 1-2y & 0\end{array}\right|=0-0=0$

c)

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

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

Egalam termenii si obtinem:

x+2y=x+1

2y=1

$y=\frac{1}{2}$

-1=2x-2

2x=1

$x=\frac{1}{2}$

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

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