Question

Se consideră matricele $A=\left(\begin{array}{rr}3 & -2 \\ 5 & -3\end{array}\right), I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ şi $B(x)=A-x I_{2}$, unde $x$ este număr real.

$5 p$ a) Arătați că det $(B(0))=1$.

$5 \mathbf{p}$ b) Arătați că $A \cdot A+I_{2}=O_{2}$, unde $O_{2}=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$.

$5 p$ c) Demonstrați că $\operatorname{det}(B(x)) \geq 1 [tex], pentru orice număr real [tex] x$

Answer 1

$A=\left(\begin{array}{rr}3 & -2 \\ 5 & -3\end{array}\right)$

a)

Facem diferenta dintre produsul diagonalelor

$det(B(0))=detA=\left|\begin{array}{rr}3 & -2 \\ 5 & -3\end{array}\right|=-9-(-10)=-9+10=1$

b)

$A\cdot A+I_2=\left(\begin{array}{rr}3 & -2 \\ 5 & -3\end{array}\right)\cdot \left(\begin{array}{rr}3 & -2 \\ 5 & -3\end{array}\right)+\left(\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{rr}-1 & 0 \\ 0 & -1\end{array}\right)+\left(\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{rr}0 & 0 \\ 0 & 0\end{array}\right)=O_2$

c)

det(B(x))≥1

$det(B(x))=\left|\begin{array}{rr}3-x & -2 \\ 5 & -3-x\end{array}\right|=(3-x)(-3-x)-(-10)=-9-3x+3x+x^2+10=x^2+1\\\\x^2\geq 0\\\\x^2+1\geq 1\\\\det(B(x))\geq 1$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928466

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