Question

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

5p 1. Arătați că det $A=0$.

5p 2. Calculați $\operatorname{det}(A+B)$.

5p 3. Arătați că $A \cdot A=A$.

5p 4. Calculați $\operatorname{det}(A \cdot B-B \cdot A)$.

5p 5. Determinați numerele reale $x$ pentru care $\operatorname{det}\left(B \cdot B+x I_{2}\right)=0$.

5p 6. Determinați numerele reale $p$ şi $q$, ştiind că $(A+B)(A+B)=p A+q B+B \cdot A$.

Answer 1

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

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

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=2×(-1)-1×(-2)=-2+2=0

2)

det(A+B)=4×(-7)-4×(-6)=-28+24=-4

3)

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

4)

det(AB-BA)

$A\cdot B=\left(\begin{array}{cc}2 & 1 \\ -2 & -1\end{array}\right)\cdot \left(\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right)=\left(\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right)=O_2$

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

$AB-BA=\left(\begin{array}{cc}0 & 0 \\ 0 &0\end{array}\right)-\left(\begin{array}{cc}-2 & -1 \\ 4 & 2\end{array}\right)=\left(\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right)$

det(AB-BA)=-4+4=0

5)

$det(B\cdot B+xI_2)=0$

$B\cdot B =\left(\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right)\cdot \left(\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right)=\left(\begin{array}{cc}-8 & -12 \\ 16& 24\end{array}\right)=-4B\\\\\left(\begin{array}{cc}-8 & -12 \\ 16& 24\end{array}\right)+\left(\begin{array}{cc}x &0\\ 0& x\end{array}\right)=\left(\begin{array}{cc}-8+x & -12 \\ 16& 24+x\end{array}\right)$

$\left|\begin{array}{cc}-8+x & -12 \\ 16& 24+x\end{array}\right|=(-8+x)(24+x)+192\\\\-192-8x+24x+x^2+192=0\\\\x^2+16x=0\\\\x(x+16)=0\\\\x=0\ si\ x=-16$

6)

(A+B)(A+B)=A·A+AB+BA+B·B=A+O₂+BA-4B

A+O₂+BA-4B=pA+qB+BA

A-4B=pA+qB

p=1 si q=-4

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919024

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