Question

Se consideră matricele $A=\left(\begin{array}{lll}3 & 2 & 1 \\ 6 & 4 & 2 \\ 9 & 6 & 3\end{array}\right), I_{3}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$ şi $B=I_{3}+A$.

5p a) Arătaţi că det $A=0$.

$5 \mathbf{p}$ b) Arătaţi că matricea $I_{3}-\frac{1}{11} A$ este inversa matricei $B$.

5p c) Dați exemplu de trei matrice $U, V, T \in \mathcal{M}_{3}(\mathbb{R})$, de rang 1 , astfel încât $U+V+T=B$.

Answer 1

$A=\left(\begin{array}{lll}3 & 2 & 1 \\ 6 & 4 & 2 \\ 9 & 6 & 3\end{array}\right)$

a)

Calculam detA, adaugam primele doua linii ale determinantului si obtinem:

$det(A)=\left|\begin{array}{lll}3 & 2 & 1 \\ 6 & 4 & 2 \\ 9 & 6 & 3\end{array}\right|$

                 3    2   1

                 6   4   2

detA=(36+36+36)-(36+36+36)=108-108=0

b)

$B=\left(\begin{array}{lll}4 & 2 & 1 \\ 6 & 5 & 2 \\ 9 & 6 & 4\end{array}\right)$

$detB=\left|\begin{array}{lll}4 & 2 & 1 \\ 6 & 5 & 2 \\ 9 & 6 & 4\end{array}\right|$

              4    2   1

              6    5   2

detB=80+36+36-(45+48+48)=152-141=11

Transpusa matricei B si inversa sa

$B^t=\left(\begin{array}{lll}4 & 6 &9 \\ 2 & 5 & 6 \\ 1 & 2 & 4\end{array}\right)$

$B^*=\left(\begin{array}{lll}8 & -2 & -1 \\ -6 & 7 & -2 \\ -9 & -6 & 8\end{array}\right)\\\\B^{-1}=\frac{1}{detB} \cdot B^*\\\\B^{-1}=\left(\begin{array}{lll}\frac{8}{11} & \frac{-2}{11} & \frac{-1}{11} \\ \frac{-6}{11} & \frac{7}{11} & \frac{-2}{11} \\ \frac{-9}{11} & \frac{-6}{11} & \frac{8}{11} \end{array}\right)$

$I_3-\frac{1}{11}A=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)-\left(\begin{array}{lll}\frac{3}{11} & \frac{2}{11} & \frac{1}{11} \\ \frac{6}{11} & \frac{4}{11} & \frac{2}{11} \\ \frac{9}{11} & \frac{6}{11} & \frac{3}{11} \end{array}\right)=\left(\begin{array}{lll}\frac{8}{11} & \frac{-2}{11} & \frac{-1}{11} \\ \frac{-6}{11} & \frac{7}{11} & \frac{-2}{11} \\ \frac{-9}{11} & \frac{-6}{11} & \frac{8}{11} \end{array}\right)=B^{-1}$

c)

Luam matricea U si pastra prima linie apoi restul termenilor 0

Luam matricea V pastram linia a doua si restul termenilor 0

Luam matricea T pastram linia a treia si restul termenilor 0

Daca le adunam ne da matricea B

Rangul lui U, V si T este 0

detU=detV=detT=0 (au cel putin o linie cu termenii 0), iar minorii lor au determinantii 0

$U=\left(\begin{array}{lll}4 & 2 & 1 \\ 0&0&0 \\ 0&0&0\end{array}\right)$

$V=\left(\begin{array}{lll}0&0&0 \\ 6 & 5 & 2 \\ 0&0&0\end{array}\right)$

$T=\left(\begin{array}{lll}0&0&0 \\ 0&0&0 \\ 9 & 6 & 4\end{array}\right)$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919124

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