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

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

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

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

[tex]A=\left(\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right)[/tex]

[tex]B(x)=\left(\begin{array}{cc}-1 & x \\ x-1 & -1\end{array}\right)[/tex]

a)

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

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

b)

[tex]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[/tex]

c)

[tex]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)[/tex]

x-2=0

x=2

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

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