Răspuns :
[tex]I_{3}=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)[/tex]
[tex]A(a)=\left(\begin{array}{ccc}a & 0 & 2-a \\ 0 & 2 & 0 \\ 2-a & 0 & a\end{array}\right)[/tex]
a)
Calculam A(2), inlocuind pe a cu 2 si obtinem:
[tex]A(2)=\left(\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right)[/tex]
Calculam det(A(2)), adaugand primele doua linii:
[tex]det(A(2))=\left|\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right|[/tex]
2 0 0
0 2 0
det(A(2))=(8+0+0)-(0+0+0)=8
b)
[tex]A(a)=\left(\begin{array}{ccc}a & 0 & 2-a \\ 0 & 2 & 0 \\ 2-a & 0 & a\end{array}\right)[/tex]
[tex]A(b)=\left(\begin{array}{ccc}a & 0 & 2-b \\ 0 & 2 & 0 \\ 2-b & 0 & b\end{array}\right)[/tex]
[tex]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)[/tex]
[tex]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)=[/tex]
[tex]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)[/tex]
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)
[tex]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)[/tex]
Dam factor comun pe 2 si obtinem:
[tex]A(a)\times A(b)=2A(ab-a-b+2)[/tex]
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
[tex]2I_3=\left(\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right)=A(2)[/tex]
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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