Question

Se considera matricea A(x)= 1-x.  0.  2x
                                                      0.    1.   0
                                                       -x.  0.   1+2x , unde x numar real .
5p a) Arătați că det (A(1))=2
5p b) Arătați că A(x) A(y)=A( xy+ x +y ), pentru orice numere reale x și y .
5p c) Determinați numerele reale x , știind că A(x)A(x) A(x) =A(7)

Answer 1

$\displaystyle A(x)= \left(\begin{array}{ccc}1-x&0&2x\\0&1&0\\-x&0&1+2x\end{array}\right) \\ \\ a).det(A(1))= \left|\begin{array}{ccc}1-1&0&2 \cdot 1\\0&1&0\\-1&0&1+2 \cdot 1\end{array}\right|= \left|\begin{array}{ccc}0&0&2\\0&1&0\\-1&0&3\end{array}\right|= \\ \\ =0 \cdot 1 \cdot 3+2 \cdot 0 \cdot 0+0 \cdot 0 \cdot (-1)-2 \cdot 1 \cdot (-1)-0 \cdot 0 \cdot 3-0 \cdot 0 \cdot 0= \\ \\ =0+0+0+2-0-0=2 \Rightarrow det(A(1))=2$

$\displaystyle b).A(x) \cdot A(y)=A(xy+x+y) \\ \\ \left(\begin{array}{ccc}1-x&0&2x\\0&1&0\\-x&0&1+2x\end{array}\right) \cdot \left(\begin{array}{ccc}1-y&0&2y\\0&1&0\\-y&0&1+2y\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}-xy-x-y+1&0&2xy+2x+2y\\0&1&0\\-x-xy-y&0&2xy+2x+2y+1\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}1-(xy+x+y)&0&2(xy+x+y)\\0&1&0\\-(xy+x+y)&0&2(xy+x+y)+1\end{array}\right)=A(xy+x+y) \Rightarrow \\ \\ \Rightarrow A(x) \cdot A(y)=A(xy+x+y)$

$c).A(x) \cdot A(x) \cdot A(x)=A(7) \\ \\ A^3=A^2 \cdot A \\ \\ A^2= \left(\begin{array}{ccc}1-x&0&2x\\0&1&0\\-x&0&1+2x\end{array}\right) \cdot \left(\begin{array}{ccc}1-x&0&2x\\0&1&0\\-x&0&1+2x\end{array}\right) = \\ \\ = \left(\begin{array}{ccc}-x^2-2x+1&0&2x^2+4x\\0&1&0\\-x^2-2x&0&2x^2+4x+1\end{array}\right) \\ \\ A^3=\left(\begin{array}{ccc}-x^2-2x+1&0&2x^2+4x\\0&1&0\\-x^2-2x&0&2x^2+4x+1\end{array}\right) \cdot \left(\begin{array}{ccc}1-x&0&2x\\0&1&0\\-x&0&1+2x\end{array}\right) =$
$\displaystyle = \left(\begin{array}{ccc}-x^3-3x^2-3x+1&0&2x ^3+6x^2+6x\\0&1&0\\-x^3-3x^2-3x&0&2x^3+6x^2+6x+1\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}1-(x^3+3x^2+3x)&0&2(x^3+3x^2+3x)\\0&1&0\\-(x^3+3x^2+3x)&0&2(x^3+3x^2+3x)+1\end{array}\right)= \\ \\ =A(x^3+3x^2+3x) \\ \\ A(x^3+3x^2+3x)=A(7) \\ \\ x^3+3x^2+3x=7 \Rightarrow x=1$