In Matrices, Is (A^(−1) )B = B(A^(−1) ) ?


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Question Number 41332 by rahul 19 last updated on 05/Aug/18

$In Matrices,$ $Is (A^{- 1}) B = B (A^{- 1}) ?$
Commented by rahul 19 last updated on 06/Aug/18

$ok prof .$
Commented by candre last updated on 06/Aug/18

$\| \begin{matrix} 1 \\ 2 \end{matrix} \| \| \begin{matrix} 1 & 2 \end{matrix} \| = \| \begin{matrix} 1 \\ 2 \end{matrix} \|$ $\| \begin{matrix} 1 & 2 \end{matrix} \| \| \begin{matrix} 1 \\ 2 \end{matrix} \| = \| \begin{matrix} 3 \end{matrix} \|$ $also AB e x i s t i n g {d o n}^{'} t e v e n i m p l y B A e x i s t$ $so, in general A B \neq B A$ $B u t i t n o t n e c e s s a r y m e a n t h a t t h e r e n o v a l u e s$ $s u c h A B = B A, f o r s q u a r e s m a t r i z e s$ $I A = A I$ $w h e r e I i s i n d e n t y m a t r i z$
Commented by maxmathsup by imad last updated on 05/Aug/18

$g e n e r a l l y t w o m a t r i c e d o n t c o m m u t e m e a n s A . B \neq B . A s o$ $g e n a r a l l y A^{- 1} . B \neq B . A^{- 1}$
	Answered by candre last updated on 05/Aug/18

	$no, matriz multiplication {don}^{'} t commute in general .$
	Commented by rahul 19 last updated on 05/Aug/18

	$ok sir .$
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