Question

Se consideră matricele $A=\left(\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right), B(x)=\left(\begin{array}{ll}1 & 0 \\ x & 1\end{array}\right)$ şi $C(x)=\left(\begin{array}{ll}1 & x \\ 2 & 3\end{array}\right)$, unde $x$ este număr real.

5p 1. Arătați că $\operatorname{det} A=3$.

5p 2. Determinați numărul real $x$ pentru care $C(x) \cdot B(x)=A$.

5p 3. Arătați că [tex]$C(x) \cdot B(x)-B(x) \cdot C(x)=\left(\begin{array}{cc}x^{2} & 0 \\ 2 x & -x^{2}\end{array}\right)$[tex], pentru orice număr real [tex][tex][tex][tex][tex]$x$[tex][tex][tex][tex][tex].

5p 4. Pentru $x=0$, determinați matricea $X \in \mathcal{M}_{2}(\mathbb{R})$ pentru care $X \cdot B(x)=A \cdot C(x)$.

5p 5. Demonstrați că, pentru orice număr întreg $x$, matricea $C(x)$ este inversabilă.

5p 6. Determinatii numerele naturale $x$ pentru care det $(B(x)+C(x))\ \textgreater \ 0$.

Answer 1

$A=\left(\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right)$

$B(x)=\left(\begin{array}{ll}1 & 0 \\ x & 1\end{array}\right)$

$C(x)=\left(\begin{array}{ll}1 & x \\ 2 & 3\end{array}\right)$

1)

Aratati ca detA=3

Facem diferenta dintre produsul diagonalelor

detA=3-0=3

2)

C(x)·B(x)=A

$C(x)\cdot B(x)=\left(\begin{array}{ll}1 & x \\ 2 & 3\end{array}\right)\cdot \left(\begin{array}{ll}1 & 0 \\ x & 1\end{array}\right)=\left(\begin{array}{ll}1 +x^2& x \\ 2+3x & 3\end{array}\right)$

$\left(\begin{array}{ll}1 +x^2& x \\ 2+3x & 3\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right)$

x=0

3)

$C(x)\cdot B(x)=\left(\begin{array}{ll}1 & x \\ 2 & 3\end{array}\right)\cdot \left(\begin{array}{ll}1 & 0 \\ x & 1\end{array}\right)=\left(\begin{array}{ll}1 +x^2& x \\ 2+3x & 3\end{array}\right)$

$B(x)\cdot C(x)=\left(\begin{array}{ll}1 & 0 \\ x & 1\end{array}\right)\cdot \left(\begin{array}{ll}1 & x \\ 2 & 3\end{array}\right)=\left(\begin{array}{ll}1& x \\ x+2& x^2+3\end{array}\right)$

$C(x)\cdot B(x)-B(x)\cdot C(x)=\left(\begin{array}{ll}1 +x^2& x \\ 2+3x & 3\end{array}\right)-\left(\begin{array}{ll}1 & x \\ x+2 & x^2+3\end{array}\right)=\left(\begin{array}{ll}x^2&0 \\ 2x & -x^2\end{array}\right)$

4)

x=0

X·B(x)=A·C(x)

X·B(0)=A·C(0)

Observam ca B(0)=I₂

⇒X=A·C(0)

$X=A\cdot C(0)=\left(\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right)\cdot \left(\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 8 & 9\end{array}\right)$

5)

Matricea C(x) este inversabila daca det(C(x)) este diferit de zero

det(C(x))=3-2x

3-2x=0

3=2x

$x=\frac{3}{2}\notin Z$⇒ 3-2x≠0⇒ matricea C(x) este inversabila

6)

det(B(x)+C(x))>0

$B(x)+C(x)=\left(\begin{array}{ll}1 & 0 \\ x & 1\end{array}\right)+ \left(\begin{array}{ll}1 & x \\ 2 & 3\end{array}\right)=\left(\begin{array}{ll}2& x \\ x+2& 4\end{array}\right)$

$\left|\begin{array}{ll}2& x \\ x+2& 4\end{array}\right|=8-x^2-2x$

-x²-2x+8>0

Δ=4+32=36

$x_1=\frac{2+6}{-2} =-4\\\\x_2=\frac{2-6}{-2} =2$

x                   -∞        -4        2         +∞

-x²-2x+8      - - - - - - 0 + + +0 - - - - -

x∈(-4,2), dar x∈N⇒ x={0,1}

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9905506

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