Question

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

5p 1. Arătați că det $A=-6$.

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

5p 3. Arătaţi că $A \cdot A-4 A=6 I_{2}$.

$5 p$ 4. Determinaţi numerele reale $x$, știind că $\operatorname{det}\left(A-x I_{2}\right)=-1$.

5 5. Determinaţi numărul real $a$, ştiind că $A \cdot A \cdot A=a A+24 I_{2}$.

5p 6. Determinați numerele reale $a$ şi $b$ pentru care $A \cdot X=X \cdot A$, unde $X=\left(\begin{array}{ll}2 & 1 \\ a & b\end{array}\right)$.

Answer 1

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

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=3-9=-6

2)

$A\cdot B=\left(\begin{array}{ll}1 & 3 \\ 3 & 3\end{array}\right)\cdot \left(\begin{array}{ll}-\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{6} \end{array}\right)=\left(\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right)=I_2$

3)

$A\cdot A=\left(\begin{array}{ll}1 & 3 \\ 3 & 3\end{array}\right)\cdot \left(\begin{array}{ll}1 & 3 \\ 3 & 3\end{array}\right)=\left(\begin{array}{ll}10 & 12 \\ 12 & 18\end{array}\right)\\\\\left(\begin{array}{ll}10 & 12 \\ 12 & 18\end{array}\right)-\left(\begin{array}{ll}4& 12 \\ 12 & 12\end{array}\right)=\left(\begin{array}{ll}6 &0 \\ 0&6\end{array}\right)=6I_2$

4)

det(A+xI₂)=-1

(1-x)(3-x)-9=-1

3-4x+x²-9+1=0

x²-4x-5=0

Δ=16+20=36

$x_1=\frac{4-6}{2} =-1\\\\x_2=\frac{4+6}{2} =5$

5)

Din punctul 3 avem :

A·A-4A=6I₂

A·A=4A+6I₂

A·A·A=(4A+6I₂)·A=4A·A+6I₂·A=4A·A+6A=4(4A+6I₂)+6A=16A+24I₂+6A=22A+24I₂

22A+24I₂=aA+24I₂

22A=aA

a=22

6)

$A\cdot X=\left(\begin{array}{ll}1 & 3 \\ 3 & 3\end{array}\right)\cdot \left(\begin{array}{ll}2& 1 \\ a & b\end{array}\right)=\left(\begin{array}{ll}2+3a & 1+3b \\ 6+3a & 3+3b\end{array}\right)$

$X\cdot A=\left(\begin{array}{ll}2& 1 \\ a & b\end{array}\right)\cdot \left(\begin{array}{ll}1 & 3 \\ 3 & 3\end{array}\right)=\left(\begin{array}{ll}5 & 9 \\ a+3b & 3a+3b\end{array}\right)$

2+3a=5

3a=3

a=1

1+3b=9

3b=8

$b=\frac{8}{3}$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919034

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