Question

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

$5 p$ a) Arătați că $\operatorname{det}(A(2))=8$.

$5 p$ b) Demonstrați că $A(a) A(b)=2 A(a b-a-b+2)$, pentru orice numere reale $a$ şi $b$.

$5 p$ c) Determinaţi perechile de numere întregi $p$ şi $q$ pentru care $A(p) A(q)=4 I_{3}$.

Answer 1

$I_{3}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$

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

a)

Calculam A(2), inlocuind pe a cu 2 si obtinem:

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

Calculam det(A(2)), adaugand primele doua linii:

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

                     2    0   0

                     0    2   0

det(A(2))=(8+0+0)-(0+0+0)=8

b)

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

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

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

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

$A(a)\times A(b)=\left(\begin{array}{ccc}ab+(2-a)(2-b) & 0 & a(2-b)+b(2-a) \\ 0 & 4 & 0 \\ b(2-a)+a(2-b) & 0 & (2-a)(2-b)+ab\end{array}\right)$

ab+(2-a)(2-b)=ab+4-2b-2a+ab=2ab-2a-2b+4=2(ab-a-b+2)

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

$A(a)\times A(b)=\left(\begin{array}{ccc}2(ab-a-b+2) & 0 & 2(-ab+a+b)\\ 0 & 4 & 0 \\ (2-ab+a+b) & 0 & 2(ab-a-b+2)\end{array}\right)$

Dam factor comun pe 2 si obtinem:

$A(a)\times A(b)=2A(ab-a-b+2)$

c)

Ne folosim de punctul b si vom avea:

A(p)A(q)=2A(pq-p-q+2)

2A(pq-p-q+2)=4I₃   |:2

A(pq-p-q+2)=2I₃

$2I_3=\left(\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right)=A(2)$

A(pq-p-q+2)=A(2)

pq-p-q+2=2

pq-p-q=0

p(q-1)-(q-1)-1=0

(q-1)(p-1)=1

Pentru ca p si q sa fie numere intregi atunci avem urmatoarele cazuri:

Cazul 1:

q-1=1 si p-1=1

q=2 si p=2

Cazul 2:

q-1=-1 si p-1=-1

q=0 si p=0

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/131774

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