Răspuns :
[tex]A=\left(\begin{array}{ll}5 & 2 \\ 2 & 1\end{array}\right)[/tex]
1)
Calculam detA, facem diferenta dintre produsul diagonalelor
detA=5-4=1
2)
[tex]A\cdot A=\left(\begin{array}{ll}5 & 2 \\ 2 & 1\end{array}\right)\cdot \left(\begin{array}{ll}5 & 2 \\ 2 & 1\end{array}\right)=\left(\begin{array}{ll}29 & 12 \\ 12 &5\end{array}\right)\\\\A\cdot A-6A=\left(\begin{array}{ll}29 & 12 \\ 12 &5\end{array}\right)-\left(\begin{array}{ll}30 & 12 \\ 12 &6\end{array}\right)=\left(\begin{array}{ll}-1 & 0 \\ 0 &-1\end{array}\right)=-I_2[/tex]
3)
det(xA)=4
5x×x-2x×2x=4
5x²-4x²=4
x²=4
x=2 si x=-2
4)
Aratati ca det(A·A-6A+aI₂)≥0
Ne folosim de punctul 2
A·A-6A+aI₂=-I₂+aI₂=(a-1)I₂
det((a-1)I₂)=(a-1)²≥0, fiind un numar la putere para
5)
m(det(A+I₂)+det(A-I₂))=det(mA)
[tex]A+I_2=\left(\begin{array}{ccc}6&2\\2&2\end{array}\right) \\\\det(A+I_2)=12-4=8\\\\\\A-I_2=\left(\begin{array}{ccc}4&2\\2&0\end{array}\right) \\\\det(A-I_2)=-4\\\\\\\\det(mA)=5m^2-4m^2=m^2\\\\m(8-4)=m^2\\\\4m-m^2=0\\\\m(4-m)=0\\\\m=0\ si \ m=4[/tex]
6)
det(mA)-det(nA)=8
Avem calculul la punctul anterior pentru det(mA)=m²
det(nA)=n²
m²-n²=8
(m-n)(m+n)=8
m si n∈Z
Caz 1:
m-n=1
m+n=8
Le adunam
2m=9
m=4,5 ∉Z Nu se poate
Caz 2:
m-n=-1
m+n=-8
Le adunam 2m=-9 Nu se poate
Caz 3:
m-n=2
m+n=4
Le adunam
2m=6
m=3 si n=1
Analog m=-3 si n=-1
Caz 4:
m-n=4
m+n=2
Le adunam
2m=6
m=3 si n=-1
Analog m=-3 si n=1
Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928526
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