Question

1. Se consideră matricea A(a)= unde a este număr real a 4 prima linie,-4 a linia 2.Arătați că det(A(-1))=17. b) Demonstrați că A(2019-a) + A(2019+a)= 2 A(2019), pentru orice număr real a. c) Determinați perechile de numere reale x și y, pentru care A(x)4(y)=2A(-8).​

Answer 1

$A(a) = \left(\begin{array}{cc}a&4\\-4&a\end{array}\right),\, a\in\mathbb{R}$

a.

$det(A(-1)) = \left|\begin{array}{cc}-1&4\\-4&-1\end{array}\right| = 1 + 16 = 17$

b.

$A(2019 - a) + A(2019 + a) =\\= \left(\begin{array}{cc}2019-a&4\\-4&2019-a\end{array}\right) + \left(\begin{array}{cc}2019+a&4\\-4&2019+a\end{array}\right)=\\\\= \left(\begin{array}{cc}2019-a+2019+a&4+4\\-4-4&2019-a+2019+a\end{array}\right)=\\\\= \left(\begin{array}{cc}2\cdot2019&2\cdot4\\2\cdot(-4)&2\cdot2019\end{array}\right) = 2\cdot \left(\begin{array}{cc}2019&4\\-4&2019\end{array}\right) = 2A(2019),\forall a\in\mathbb{R}$

c.

$A(x)A(y) = 2A(-8)\\ \left(\begin{array}{cc}x&4\\-4&x\end{array}\right)\cdot \left(\begin{array}{cc}y&4\\-4&y\end{array}\right)=2\cdot \left(\begin{array}{cc}-8&4\\-4&-8\end{array}\right)\\\\ \left(\begin{array}{cc}xy-16&4x+4y\\-4y-4x&-16+xy\end{array}\right)= \left(\begin{array}{cc}-16&8\\-8&-16\end{array}\right)\\\\\Rightarrow\displaystyle \left \{ {{xy-16=-16} \atop {4(x+y)=8}} \right. \Rightarrow \left \{ {{xy=0} \Rightarrow (x=0\text{ sau } y=0)\atop {x+y=2}} \right.$

$\Rightarrow\displaystyle \left \{ {{x=0} \atop {y=2}} \right. \text{ sau } \left \{ {{y=0} \atop {x=2}} \right. \Rightarrow S = \{(0,2),\,(2,0)\}$