exo7 3711

Reconnaître les endomorphismes de R3\R^3 définis par les expressions analytiques dans la base canonique :

1

{3x′=2x+2y+z3y′=−2x+y+2z3z′=x−2y+2z\begin{cases} 3x' = 2x+2y+z \cr 3y' = -2x+y+2z \cr 3z' = x-2y+2z \cr\end{cases}

2

{9x′=8x+y−4z9y′=−4x+4y−7z9z′=x+8y+4z\begin{cases} 9x' = 8x+y-4z \cr 9y' = -4x+4y-7z \cr 9z'=x+8y+4z \cr\end{cases}

3

{3x′=−2x+2y−z3y′=2x+y−2z3z′=−x−2y−2z\begin{cases} 3x' = -2x+2y-z \cr 3y' = 2x+y-2z \cr 3z' = -x-2y-2z \cr\end{cases}

4

{4x′=−2x−y6+z64y′=x6+y+3z4z′=−x6+3y+z\begin{cases} 4x' = -2x - y\sqrt6 + z\sqrt6 \cr 4y' = x\sqrt6 + y + 3z \cr 4z' = -x\sqrt6 + 3y + z \cr \end{cases}

5

{x′=x3+y2−z6y′=x3+2z6z′=x3−y2−z6\begin{cases}x' = \frac x{\sqrt3} + \frac y{\sqrt2} - \frac z{\sqrt6} \cr y' = \frac x{\sqrt3} + \frac {2z}{\sqrt6} \cr z' = \frac x{\sqrt3} - \frac y{\sqrt2} - \frac z{\sqrt6} \cr\end{cases}

6

{3x′=x+2y+2z3y′=2x+y−2z3z′=2x−2y+z\begin{cases} 3x' = x+2y+2z \cr 3y' = 2x+y-2z \cr 3z' = 2x-2y+z \cr\end{cases}

7

{7x′=−2x+6y−3z7y′=6x+3y+2z7z′=−3x+2y+6z\begin{cases} 7x' = -2x+6y-3z \cr 7y' = 6x+3y+2z \cr 7z' = -3x+2y+6z \cr\end{cases}

8

{3x′=2x−2y+z3y′=−2x−y+2z3z′=x+2y+2z\begin{cases} 3x' = 2x-2y+z \cr 3y' = -2x-y+2z \cr 3z' = x+2y+2z \cr\end{cases}

9

{3x′=2x+y+2z3y′=2x−2y−z3z′=−x−2y+2z\begin{cases} 3x' = 2x+y+2z \cr 3y' = 2x-2y-z \cr 3z' = -x-2y+2z \cr\end{cases}

10

{4x′=−x+3y−z64y′=3x−y−z64z′=x6+y6+2z\begin{cases} 4x' = -x+3y-z\sqrt6 \cr 4y' = 3x-y-z\sqrt6 \cr 4z' = x\sqrt6+y\sqrt6+2z \cr\end{cases}

11

{15x′=5x−10z15y′=−8x+5y+6z15z′=6x−10y+8z\begin{cases} 15x'=5x-10z \cr 15y' = -8x+5y+6z \cr 15z' = 6x-10y+8z \cr\end{cases}

Exercice yX4w

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