Question

Se consideră matricele $O_{3}=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right), I_{3}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$ şi $A(a)=\left(\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1\end{array}\right)$, unde $a$ este număr real.

5p a) Arătaţi că det $(A(a))=1$, pentru orice număr real $a$.

$5 p$ b) Se consideră matricea $B(a)=A(a)-I_{3}$, unde $a$ este număr real. Demonstraţi că, pentru orice număr real $a, B(a) \cdot B(a) \cdot B(a)=O_{3}$.

$5 p$ c) Determinați numărul natural nenul $n$, știind că suma elementelor matricei $X$ pentru care $A(2) \cdot X=A(1)+A(2)+\ldots+A(n)$ este egală cu 21 .

Answer 1

$A(a)=\left(\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1\end{array}\right)$

a)

Calculam det(A(1)), inlocuim pe a cu 1 si adaugam primele doua linii ale determinantului

$det(A(1))=\left|\begin{array}{lll}1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1\end{array}\right|$

                      1    a   0

                      0   1    a

det(A(1))=(1+0+0)-(0+0+0)=1

b)

$B(a)=\left(\begin{array}{lll}0& a & 0 \\ 0 & 0 & a \\ 0 & 0 & 0\end{array}\right)$

$B(a)\cdot B(a)=\left(\begin{array}{lll}0& a & 0 \\ 0 & 0 & a \\ 0 & 0 & 0\end{array}\right)\cdot \left(\begin{array}{lll}0& a & 0 \\ 0 & 0 & a \\ 0 & 0 & 0\end{array}\right)=\left(\begin{array}{lll}0& 0 & a^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$

$B(a)\cdot B(a)\cdot B(a)=\left(\begin{array}{lll}0& a & 0 \\ 0 & 0 & a \\ 0 & 0 & 0\end{array}\right)\cdot \left(\begin{array}{lll}0& 0 & a^2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)=\left(\begin{array}{lll}0& 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)=O_3$

c)

$Fie\ M=A(1)+A(2)+...+A(n)=\left(\begin{array}{lll}n & \frac{n(n+1)}{2} & 0 \\ 0 & n & \frac{n(n+1)}{2} \\ 0 & 0 & n\end{array}\right)$

$X=A(2)^{-1}\cdot M$

Calculam det(A(2))

det(A(2))=1

$A(2)^t=\left(\begin{array}{lll}1 & 0& 0 \\ 2& 1 & 0 \\ 0 &2 & 1\end{array}\right)$

$A(2)^*=\left(\begin{array}{lll}1 & -2& 4 \\ 0& 1 & -2 \\ 0 &0 & 1\end{array}\right)$

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

$X=\left(\begin{array}{lll}1 & -2& 4 \\ 0& 1 & -2 \\ 0 &0 & 1\end{array}\right)\cdot \left(\begin{array}{lll}n & \frac{n(n+1)}{2} & 0 \\ 0 & n & \frac{n(n+1)}{2} \\ 0 & 0 & n\end{array}\right)=\left(\begin{array}{lll}n & \frac{n(n+1)}{2} -2n& -n(n+1) +4n\\ 0& n & \frac{n(n+1)}{2} -2n \\ 0 &0 & n\end{array}\right)$

Suma elementelor matricei X este egala cu 21

3n=21

n=7

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919076

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