Question

Se consideră matricele $A=\left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right), B=\left(\begin{array}{cc}6 & -4 \\ -3 & 3\end{array}\right)$ şi $I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$.

5p 1. Arătați că det $A=0$.

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

5p 3. Demonstrați că $\operatorname{det}\left(A \cdot B-I_{2}\right)=\operatorname{det}\left(B \cdot A-I_{2}\right)$.

$5 p /tex] 4. Determinați numărul real [tex] x$, știiind că $B-A+x I_{2}=\left(\begin{array}{cc}2 & -2 \\ -2 & 0\end{array}\right)$.

5p 5. Demonstrați că $\operatorname{det}\left(I_{2}+a A\right)+\operatorname{det}\left(I_{2}-a A\right)=2$, pentru orice număr real $a$.

5p 6. Rezolvaţi in $\mathcal{M}_{2}(\mathbb{R})$ ecuaţia $\left(I_{2}-A\right) \cdot X=A$.

Answer 1

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

$B=\left(\begin{array}{cc}6 & -4 \\ -3 & 3\end{array}\right)$

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=2-2=0

2)

Calculam A·A-B

$A\cdot A=\left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right)=\left(\begin{array}{cc}6 & -6 \\ -3 & 3\end{array}\right)\\\\A\cdot A-B=\left(\begin{array}{cc}6 & -6 \\ -3 & 3\end{array}\right)-\left(\begin{array}{cc}6 & -4 \\ -3 & 3\end{array}\right)=\left(\begin{array}{cc}0 & -2 \\ 0 & 0\end{array}\right)$

3)

Aratati ca det(AB-I₂)=det(BA-I₂)

$A\cdot B=\left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}6& -4 \\ -3 & 3\end{array}\right)=\left(\begin{array}{cc}18 & -14 \\ -9 & 7\end{array}\right)\\\\\left(\begin{array}{cc}18 & -14 \\ -9 & 7\end{array}\right)-\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)=\left(\begin{array}{cc}17 & -14 \\ -9 & 6\end{array}\right)\\\\\left|\begin{array}{cc}17 & -14 \\ -9 & 6\end{array}\right|=102-126=-24$

$B\cdot A=\left(\begin{array}{cc}6& -4 \\ -3 & 3\end{array}\right)\cdot \left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right)=\left(\begin{array}{cc}16 & -16 \\ -9 & 9\end{array}\right)\\\\\left(\begin{array}{cc}16 & -16 \\ -9 & 9\end{array}\right)-\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)=\left(\begin{array}{cc}15 & -16 \\ -9 & 8\end{array}\right)\\\\\left|\begin{array}{cc}15 & -16 \\ -9 & 8\end{array}\right|=120-144=-24$

Din cele doua rezulta ca det(AB-I₂)=det(BA-I₂)

4)

$B-A=\left(\begin{array}{cc}6& -4 \\ -3 & 3\end{array}\right)- \left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right)=\left(\begin{array}{cc}4 & -2 \\ -2 & 2\end{array}\right)\\\\\left(\begin{array}{cc}4 & -2 \\ -2 & 2\end{array}\right)+ \left(\begin{array}{cc}x &0 \\ 0 & x\end{array}\right)=\left(\begin{array}{cc}2& -2 \\ -2 & 0\end{array}\right)$

4+x=2

x=-2

5)

det(I₂+aA)+det(I₂-aA)=2

$\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)+\left(\begin{array}{cc}2a & -2a \\ -a & a\end{array}\right)=\left(\begin{array}{cc}1+2a & -2a \\ -a & 1+a\end{array}\right)$

$\left|\begin{array}{cc}1+2a & -2a \\ -a & 1+a\end{array}\right|=(1+2a)(1+a)-2a^2=1+3a+2a^2-2a^2=1+3a$

$\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)-\left(\begin{array}{cc}2a & -2a \\ -a & a\end{array}\right)=\left(\begin{array}{cc}1-2a & 2a \\ a & 1-a\end{array}\right)$

$\left|\begin{array}{cc}1-2a & 2a \\ a & 1-a\end{array}\right|=1-3a+2a^2-2a^2=1-3a$

det(I₂+aA)+det(I₂-aA)=1+3a+1-3a=2

6)

Notam $C=I_2-A$

$C=\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)-\left(\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right)=\left(\begin{array}{cc}-1 & 2 \\ 1 & 0\end{array}\right)$

$X=C^{-1}\cdot A$

Calculam inversa matricei C

$C^t=\left(\begin{array}{cc}-1 & 1 \\ 2& 0\end{array}\right)$

$C^*=\left(\begin{array}{cc}0& -2 \\ -1 & -1\end{array}\right)$

detC=0-2=-2

$C^{-1}=\left(\begin{array}{cc}0& 1 \\ \frac{1}{2} & \frac{1}{2} \end{array}\right)$

$X=\left(\begin{array}{cc}0& 1 \\ \frac{1}{2} & \frac{1}{2} \end{array}\right)\cdot \left(\begin{array}{cc}2& -2 \\ -1 & 1 \end{array}\right)=\left(\begin{array}{cc}-1& 1 \\ \frac{1}{2} &- \frac{1}{2} \end{array}\right)$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/1857848

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