Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 159327 by SANOGO last updated on 15/Nov/21

	Answered by Derrick last updated on 16/Nov/21

	$1) m o n t r o n s q u e f e s t l i n e a i r e$ $s o i e n t X = (x . y . z) e t Y = (x^{'} . y^{'} . z^{'})$ $o n a :$ $f (X + Y) = [(x + x^{'}) + 2 (y + y^{'}) + 3 (z + z^{'}); 2 (x + x^{'}) + (y + y^{'}) + 3 (z + z^{'}); - (x + x^{'}) - (y + y^{'}) - (z + z^{'})]$ $= [x + 2 y + 3 z; 2 x + y + 3 z; - x - y - z] + [x^{'} + 2 y^{'} + 3 z^{'}; 2 x^{'} + y^{'} + 3 z^{'}; - x^{'} - y^{'} - z^{'}]$ $= f (X) + f (Y)$ $a i n s i f e s t l i n e a i r e$
	Answered by Derrick last updated on 16/Nov/21
	$2)donnons la matrice de f dans la base canonique de R^3 (((1 2 1)),( { (( 2 1 3)),((−1 −1 −1 )) :}) )$
	$2) d o n n o n s l a m a t r i c e d e f d a n s l a b a s e c a n o n i q u e d e R^{3}$ $(\begin{matrix} 1 2 1 \\ {\begin{cases} 2 1 3 \\ - 1 - 1 - 1 \end{cases} \end{matrix})$
	Commented by SANOGO last updated on 16/Nov/21

	$m e r c i b i e n$
	Answered by Derrick last updated on 16/Nov/21

	$3) a) d e t e r m i n o n s l a d i m e n s i o n e t u n e b a s e d e k e r f$ $s o i t X = (x . y . z) \in R^{3}$ $X \in k e r f \Leftrightarrow A X = 0$ $\Leftrightarrow l_{1} ∙ x + 2 y + z = 0$ $l_{2} ∙ 2 x + y + 3 z = 0$ $l_{3} ∙ - x - y - z = 0$ $l_{3} \Rightarrow x = - y - z$ $l_{3} d a n s l_{2} \Rightarrow y = z$ $l_{2} e t l_{3} d a n s l_{1} \Rightarrow x = - 3 z$ $d^{'} o u : X = (x . y . z) = (- 3 z, z, z) = z (- 3, 1, 1); z \in R$ $k e r f =< (- 3, 1, 1) >= b a s e d e k e r f$ $o n d e d u i t d o n c q u e :$ $d i m (k e r f) = 1$
	Answered by Derrick last updated on 16/Nov/21

	$b) p u i s d i m (k e r f) = 1 ≢ 0 a l o r s f n^{'} e s t p a s i n j e c t i v e$ $4) a) d o n n o n s l e r a n g d e f e t u n e b a s e d e i m f$ $d^{'} a p r e s l e t h e o r e m e d u r a n g o n a :$ $d i m (k e r f) + r a n g (f) = 3$ $\Rightarrow r a n g (f) = 3 - d i m (k e r f)$ $= 3 - 1$ $= 2$ $* d o n n o n s u n e b a s e d u i m f$ $V = (1, 2, 1) e t U = (- 1, - 1, - 1) s o n t d a n s i m f e t d e p l u s i l f o r m e u n s y s t e m e l i b r e$ $d o n c u n e b a s e d u i m f e s t :$ $B =< V, U >$ $b) {r e m a r q u}^{'} o n q u e l a d i m e n s i o n d e l^{'} e s p a c e d e d e p a r t e t d^{'} a r r i v e e e s t 3$ $a i n s i d o n c :$ $f i n j e c r i v e \Leftrightarrow f s u r j e c t i v e \Leftrightarrow f b i j e c r i v e$ $p u i s q u e f n^{'} e s t p a s i n n e c t i v e a l o r s o n e n d e d u i t q u e f n^{'} e s t p a s s u r j e c t i v e$ $\dots d e r r i c k \dots$
	Commented by SANOGO last updated on 16/Nov/21

	$m e r c i b i e n$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum