Question

Se consideră matricele $A(a)=\left(\begin{array}{ll}1 & 2 \\ 3 & a\end{array}\right)$ şi $B=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)$, unde $a$ este număr real.

1. Arătaţi că $\operatorname{det}(A(1))=-5$.

2. Determinaţi numerele reale $a$, ştiind că $\operatorname{det}(a A(a))=0$.

3. Arătaţi că $\operatorname{det}(A(a) \cdot B-B \cdot A(a))=-9$, pentru orice număr real $a$

4. Demonstrați că $A(a-1)+A(a+1)=2 A(a) [tex], pentru orice număr real [tex] a$.

5. Determinaţi numărul real $a$, știind că $\operatorname{det}(A(a)+B)=a$.

6. Determinaţi numărul natural nenul $n$ pentru care $A(1)+A(2)+\ldots+A(n)=11 A(6)$.

Answer 1

$A(a)=\left(\begin{array}{ll}1 & 2 \\ 3 & a\end{array}\right)$

$B=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)$

1)

Calculam det(A(1)), inlocuim pe a cu 1 si facem diferenta dintre produsul diagonalelor

det(A(1))=1×1-2×3=1-6=-5

2)

det(A(a))=0

a-6=0

a=6

3)

det(A(a)·B-B·A(a))=-9

$A(a)\cdot B=\left(\begin{array}{ll}1 & 2 \\ 3 & a\end{array}\right)\cdot \left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 3 \\ 3 &3+a\end{array}\right)$

$B\cdot A(a)= \left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\cdot \left(\begin{array}{ll}1 & 2 \\ 3 & a\end{array}\right)=\left(\begin{array}{ll}4 & 2+a \\ 3 &a\end{array}\right)$

$A(a)\cdot B-B\cdot A(a)=\left(\begin{array}{ll}1 & 3 \\ 3 &3+a\end{array}\right)-\left(\begin{array}{ll}4 & 2+a \\ 3 &a\end{array}\right)=\left(\begin{array}{ll}-3 & 1-a \\ 0 &3\end{array}\right)$

$\left|\begin{array}{ll}-3 & 1-a \\ 0 &3\end{array}\right|=-9-0=-9$

4)

$A(a-1)+A(a+1)=\left(\begin{array}{ll}1 & 2 \\ 3 & a-1\end{array}\right)+\left(\begin{array}{ll}1 & 2 \\ 3 & a+1\end{array}\right)=\left(\begin{array}{ll}2 & 4 \\ 6 & 2a\end{array}\right)=2A(a)$

5)

det(A(a)+B)=a

$A(a)+B=\left(\begin{array}{ll}1 & 2 \\ 3 & a\end{array}\right)+\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}2 & 3 \\ 3 & a+1\end{array}\right)\\\\\left|\begin{array}{ll}2 & 3 \\ 3 & a+1\end{array}\right|=2a+2-9\\\\2a-7=a\\\\a=7$

6)

$A(1)+A(2)+...+A(n)=\left(\begin{array}{ll}n & 2n \\ 3n &\frac{n(n+1)}{2} \end{array}\right)=11A(6)\\\\\left(\begin{array}{ll}n & 2n \\ 3n &\frac{n(n+1)}{2} \end{array}\right)=11\left(\begin{array}{ll}1 & 2 \\ 3 &6 \end{array}\right)$

n=11

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919130

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