👤
a fost răspuns

Se consideră matricele [tex]$A=\left(\begin{array}{cc}3 & 13 \\ -1 & -4\end{array}\right)$[/tex] şi [tex]$I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$[/tex].

5p a) Arătați că [tex]$\operatorname{det} A=\operatorname{det}\left(A+I_{2}\right)$[/tex].

[tex]$5 p$[/tex] b) Determinați numărul real [tex]$a$[/tex], știind că [tex]$A \cdot A \cdot A=a I_{2}$[/tex].

[tex]$5 p$[/tex] c) Determinați perechile [tex]$(m, n)$[/tex] de numere naturale, cu [tex]$m \neq n$[/tex], pentru care [tex]$\operatorname{det}\left(A+m I_{2}\right)=\operatorname{det}\left(A+n I_{2}\right)$[/tex].

Răspuns :

[tex]A=\left(\begin{array}{cc}3 & 13 \\ -1 & -4\end{array}\right)[/tex]

a)

det(A+I₂)=detA

[tex]det(A+I_2)=\left|\begin{array}{cc}4 & 13 \\ -1 & -3\end{array}\right|=-12+13=1\\\\detA=-12+13=1[/tex]

Observam ca sunt egale

b)

[tex]A^2=\left(\begin{array}{cc}3 & 13 \\ -1 & -4\end{array}\right)\cdot \left(\begin{array}{cc}3 & 13 \\ -1 & -4\end{array}\right)=\left(\begin{array}{cc}-4 & -13 \\ 1 & 3\end{array}\right)\\\\A^3=\left(\begin{array}{cc}-4 & -13 \\ 1 & 3\end{array}\right)\cdot \left(\begin{array}{cc}3 & 13 \\ -1 & -4\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)[/tex]

[tex]aI_2=I_2\\\\a=1[/tex]

a=1

c)

[tex]det(A+mI_2)=det(A+nI_2)[/tex]

[tex]\left|\begin{array}{cc}3+m & 13 \\ -1 & -4+m\end{array}\right|=\left|\begin{array}{cc}3+n & 13 \\ -1 & -4+n\end{array}\right|\\\\(m-4)(m+3)+13=(n-4)(n+3)+13\\\\m^2-m+1=n^2-n+1\\\\(m-n)(m+n-1)=0\\\\m=n\ NU\\\\m+n-1=0[/tex]

m+n=1

m=0 si n =1 sau m=1 si n=0

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9882190

#BAC2022

#SPJ4

Alte intrebari