Question

Se consideră matricele $A=\left(\begin{array}{cc}1 & 3 \\ 2 & 1\end{array}\right), B=\left(\begin{array}{cc}-4 & 0 \\ 0 & 4\end{array}\right)$ şi $M(x)=\left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)$, unde $x$ este număr real.

$5 p$ a) Arătați că det $A=-5$.

$5 \mathbf{p}$ b) Arătați că $\operatorname{det}(A+M(-1))=\operatorname{det} B$.

$5 p$ c) Determinați numărul real $x$ pentru care $M(x) \cdot A-A \cdot M(x)=B$.

Answer 1

$A=\left(\begin{array}{cc}1 & 3 \\ 2 & 1\end{array}\right)$

$B=\left(\begin{array}{cc}-4 & 0 \\ 0 & 4\end{array}\right)$

$M(x)=\left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)$

a)

Facem diferenta dintre produsul diagonalelor si obtinem:

detA=1-6=-5

b)

det(A+M(-1))=detB

detB=-16-0=-16

$det(A+M(-1))=\left|\begin{array}{ll}0 & 4 \\ 4 &4\end{array}\right|=0-16=-16$

Observam ca sunt egale

c)

$M(x)\cdot A=\left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)\cdot \left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)=\left(\begin{array}{ll}x+2 & 3x+1 \\ 8 & 9\end{array}\right)$

$A\cdot M(x)= \left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)\cdot \left(\begin{array}{ll}x & 1 \\ 2 & 3\end{array}\right)=\left(\begin{array}{ll}x+6 & 10\\ 2x+2 & 5\end{array}\right)$

$\left(\begin{array}{ll}x+2 & 3x+1\\ 8 & 9\end{array}\right)-\left(\begin{array}{ll}x+6 & 10\\ 2x+2 & 5\end{array}\right)=\left(\begin{array}{ll}-4 & 3x-9\\ -2x+6 & 4\end{array}\right)$

Egalam cu termenii matricei B si obtinem

3x-9=0

3x=9

x=3

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928396

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