Question

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

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

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

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

Answer 1

$A=\left(\begin{array}{cc}3 & 13 \\ -1 & -4\end{array}\right)$

a)

det(A+I₂)=detA

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

Observam ca sunt egale

b)

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

$aI_2=I_2\\\\a=1$

a=1

c)

$det(A+mI_2)=det(A+nI_2)$

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

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

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