Question

Se consideră matricele $A=\left(\begin{array}{ll}0 & 3 \\ 6 & 9\end{array}\right)$ şi $B(x)=\left(\begin{array}{cc}2 & x \\ 2+x & 4\end{array}\right)$, unde $x$ este număr real.

5p 1. Arătați că det [tex]$A=-18$/tex].

5p 2. Arătați că $A \cdot B(0)-B(0) \cdot A=6\left(\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right)$.

5p 3. Arătați că $\operatorname{det}(B(x))=(2-x)(x+4)$, pentru orice număr real $x$.

5p 4. Arătați că det $(A+B(2))\ \textless \ \operatorname{det} A+\operatorname{det}(B(2))$.

5p 5. Demonstrați că $B(x) \cdot B(y)=B(y) \cdot B(x)$ dacă şi numai dacă $x=y$.

$5 p$ 6. Determinați numărul natural nenul $n$ pentru care $B(1)+B(2)+B(3)+\ldots+B(n)=\left(\begin{array}{cc}200 & 5050 \\ 5250 & 400\end{array}\right)$.

Answer 1

$A=\left(\begin{array}{ll}0 & 3 \\ 6 & 9\end{array}\right)$

$B(x)=\left(\begin{array}{cc}2 & x \\ 2+x & 4\end{array}\right)$

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=9×0-6×3=0-18=-18

2)

Calculam A·B(0)

$A\cdot B(0)=\left(\begin{array}{ll}0 & 3 \\ 6 & 9\end{array}\right)\cdot \left(\begin{array}{ll}2 & 0 \\ 2 & 4\end{array}\right)=\left(\begin{array}{ll}6 & 12 \\ 30 & 36\end{array}\right)$

Calculam B(0)·A

$B(0)\cdot A= \left(\begin{array}{ll}2 & 0 \\ 2 & 4\end{array}\right)\cdot \left(\begin{array}{ll}0 & 3 \\ 6 & 9\end{array}\right)=\left(\begin{array}{ll}0 &6 \\24 & 42\end{array}\right)$

$\left(\begin{array}{ll}6 &12 \\30 & 36\end{array}\right)-\left(\begin{array}{ll}0 &6 \\24 & 42\end{array}\right)=\left(\begin{array}{ll}6 &6 \\6 &-6\end{array}\right)=6\left(\begin{array}{ll}1 &1 \\1 & -1\end{array}\right)$

3)

Calculam det(B(x))

det(B(x))=(2×4)-(2+x)x=8-2x-x²=-x²-2x+8

-x²-2x+8=-x²-4x+2x+8=-x(x+4)+2(x+4)=(x+4)(2-x)

4)

det(A+B(2))<detA+det(B(2))

Calculam det(A+B(2))

det(A+B(x))=2×13-(3+x)(8+x)=26-24-3x-8x-x²=-x²-11x+2

det(A+B(2))=-4-22+2=-24

detA=-1

det(B(2))=0

-24<-1+0

-24<-1

5)

$B(x)\cdot B(y)=\left(\begin{array}{cc}2 & x \\ 2+x & 4\end{array}\right)\cdot \left(\begin{array}{cc}2 & y \\ 2+y & 4\end{array}\right)=\left(\begin{array}{cc}4+2x+xy & 2y+4x \\ 12+2x+4y & 16+2y+xy\end{array}\right)$

$B(y)\cdot B(x)=\left(\begin{array}{cc}2 & y \\ 2+y & 4\end{array}\right)\cdot \left(\begin{array}{cc}2 & x \\ 2+x & 4\end{array}\right)=\left(\begin{array}{cc}4+2y+xy & 2x+4y \\ 12+2y+4x & 16+2x+xy\end{array}\right)$

4+2x+xy=4+2y+xy

2x=2y

x=y

6)

$B(1)+B(2)+...+B(n)=\left(\begin{array}{cc}2n & 1+2+...+n \\ 2n+1+2+...+n & 4n\end{array}\right)\\\\\left(\begin{array}{cc}2n & 1+2+...+n \\ 2n+1+2+...+n & 4n\end{array}\right)=\left(\begin{array}{cc}2n &\frac{n(n+1)}{2} \\ 2n+\frac{n(n+1)}{2}& 4n\end{array}\right)\\\\\left(\begin{array}{cc}2n &\frac{n(n+1)}{2} \\ 2n+\frac{n(n+1)}{2}& 4n\end{array}\right)=\left(\begin{array}{cc}200 &5050 \\ 5250& 400\end{array}\right)$

2n=200

n=100

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9918886

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