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

5p 1. Arătați că det [tex]$A=0$[/tex].

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

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

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

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

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

Răspuns :

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

[tex]B=\left(\begin{array}{cc}6 & -4 \\ -3 & 3\end{array}\right)[/tex]

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=2-2=0

2)

Calculam A·A-B

[tex]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)[/tex]

3)

Aratati ca det(AB-I₂)=det(BA-I₂)

[tex]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[/tex]

[tex]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[/tex]

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

4)

[tex]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)[/tex]

4+x=2

x=-2

5)

det(I₂+aA)+det(I₂-aA)=2

[tex]\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)[/tex]

[tex]\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[/tex]

[tex]\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)[/tex]

[tex]\left|\begin{array}{cc}1-2a & 2a \\ a & 1-a\end{array}\right|=1-3a+2a^2-2a^2=1-3a[/tex]

det(I₂+aA)+det(I₂-aA)=1+3a+1-3a=2

6)

Notam [tex]C=I_2-A[/tex]

[tex]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)[/tex]

[tex]X=C^{-1}\cdot A[/tex]

Calculam inversa matricei C

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

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

detC=0-2=-2

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

[tex]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)[/tex]

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

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