Question

Se consideră matricea $A(a)=\left(\begin{array}{ccc}1 & a & a^{2}-a \\ 0 & 1 & 2 a \\ 0 & 0 & 1\end{array}\right)$, unde $a$ este număr real.

$5 \mathbf{p}$ a) Arătaţi că $\operatorname{det}(A(1))=1$.

$5 p$ b) Demonstrați că $A(a) A(b)=A(a+b)$, pentru orice numere reale $a$ și $b$.

$5 p$ c) Determinați matricea $X \in \mathcal{M}_{3}(\mathbb{R})$ pentru care $A(3) \cdot X=A(5)$.

Answer 1

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

a) det(A(1))=1

Mai intai formam matricea A(1), inlocuind pe a cu 1

$A(1)=\left(\begin{array}{ccc}1 & 1& 1^{2}-1 \\ 0 & 1 & 2 \times1 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & 1& 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right)$

Calculam det(A(1)), adaugand primele doua linii

$detA(1)=\left|\begin{array}{ccc}1 & 1& 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right|$

                   1    1    0

                   0   1    2

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

b) A(a)×A(b)=A(a+b)

$A(a)=\left(\begin{array}{ccc}1 & a & a^{2}-a \\ 0 & 1 & 2 a \\ 0 & 0 & 1\end{array}\right)\\\\\\A(b)=\left(\begin{array}{ccc}1 & b & b^{2}-b \\ 0 & 1 & 2 b \\ 0 & 0 & 1\end{array}\right)$

$A(a+b)=\left(\begin{array}{ccc}1 & a+b & (a+b)^{2}-(a+b) \\ 0 & 1 & 2 (a+b) \\ 0 & 0 & 1\end{array}\right)$

Calculam A(a)×A(b)

$A(a)\times A(b)=\left(\begin{array}{ccc}1 & a & a^{2}-a \\ 0 & 1 & 2 a \\ 0 & 0 & 1\end{array}\right)\times \left(\begin{array}{ccc}1 & b & b^{2}-b \\ 0 & 1 & 2 b \\ 0 & 0 & 1\end{array}\right)=$

$=\left(\begin{array}{ccc}1 & b+a & b^2-b+2ab+a^2-a \\ 0 & 1 & 2b+2a \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & b+a & (a+b)^2-(a+b) \\ 0 & 1 & 2(a+b) \\ 0 & 0 & 1\end{array}\right)=A(a+b)$

Nota:

b²-b+2ab+a²-a=a²+2ab+b²-a-b=(a+b)²-(a+b)

2a+2b=2(a+b)

c) Calculam A(3) si A(5), inlocuind pe a cu 3, respectiv 5

$A(3)=\left(\begin{array}{ccc}1 & 3 & 6 \\ 0 & 1 & 6 \\ 0 & 0 & 1\end{array}\right)\\\\\\A(5)=\left(\begin{array}{ccc}1 & 5 & 20 \\ 0 & 1 & 10 \\ 0 & 0 & 1\end{array}\right)$

Ne folosim de punctul b, de faptul ca A(a)×A(b)=A(a+b)

Atunci A(3)×A(-3)=A(0)

$A(0)=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)=I_3$

Inseamna ca inversa matricei A(3) este A(-3)

A(3)×X=A(5)   |×(A-3)

Inmultim ecuatia in stanga cu A(-3) si obtinem:

A(-3)×A(3)×X=A(-3)×A(5)

A(0)×X=A(-3)×A(5)

I₃×X=A(-3)×A(5)

X=A(-3)×A(5)

X=A(-3+5)=A(2)

$X=A(2)=\left(\begin{array}{ccc}1 & 2 & 2 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{array}\right)$

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/6640879

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