Question

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

$5 \mathbf{p}$ a) Arătați că det $(B(1))=1$.

$5 p$ b) Arătați că $A \cdot A-2 A=I_{2}$.

$5 p$ c) Determinați numărul real $x$ pentru care $A \cdot B(x)=I_{2}$.

Answer 1

$A=\left(\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right)$

$B(x)=\left(\begin{array}{cc}-1 & x \\ x-1 & -1\end{array}\right)$

a)

Calculam det(B(1)), inlocuim pe x cu 1, facem diferenta dintre produsul diagonalelor

det(B(1))=1-0=1

b)

$A\cdot A=\left(\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right)\cdot \left(\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right)=\left(\begin{array}{ll}3& 4 \\ 2 & 3\end{array}\right)\\\\\left(\begin{array}{ll}3& 4 \\ 2 & 3\end{array}\right)-\left(\begin{array}{ll}2& 4 \\ 2 & 2\end{array}\right)=\left(\begin{array}{ll}1& 0 \\ 0 & 1\end{array}\right)=I_2$

c)

$A\cdot B(x)=\left(\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}-1 & x \\ x-1 & -1\end{array}\right)= \left(\begin{array}{cc}-1+2x-2 & x-2 \\ -1+x-1 & x-1\end{array}\right)\\\\\left(\begin{array}{cc}2x-3 & x-2 \\x-2 & x-1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\0& 1\end{array}\right)$

x-2=0

x=2

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9918938

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