Find an orthogonal matrix A whose first row is u


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Question Number 116746 by bemath last updated on 06/Oct/20

$Find an orthogonal matrix A whose$ $first row is u_{1} = (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) .$
	Answered by john santu last updated on 06/Oct/20

	$F i r s t s t e p f i n d a n o n z e r o v e c t o r$ $w_{1} = (x, y, z) t h a t i s o r t h o g o n a l t o u_{1} .$ $f o r w h i c h ⟨ u_{1}, w_{1} ⟩ = 0 = \frac{x}{3} + \frac{2 y}{3} + \frac{2 z}{3}$ $o r x + 2 y + 2 z = 0 . O n e s u c h s o l u t i o n$ $i s w_{1} = (0, 1, - 1) . N o r m a l i z e w_{1} t o$ $o b t a i n t h e s e c o n d r o w o f A, u_{2} = (0, \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}})$ $N e x t s t e p f i n d a n o n z e r o v e c t o r w_{2} = (x, y, z)$ $t h a t i s o r t h o g o n a l t o b o t h u_{1} a n d u_{2}$ $f o r w h i c h ⟨ u_{1}, w_{2} ⟩ = 0 a n d ⟨ u_{2}, w_{2} ⟩ = 0$ $\Rightarrow ⟨ u_{1}, w_{2} ⟩ = \frac{x}{3} + \frac{2 y}{3} + \frac{2 z}{3} = 0$ $\Rightarrow ⟨ u_{2}, w_{2} ⟩ = \frac{y}{\sqrt{2}} - \frac{z}{\sqrt{2}} = 0$ $s e t z = - 1 a n d f i n d t h e s o l u t i o n$ $w_{2} = (4, - 1, - 1) . N o r m a l i z e w_{2}$ $a n d o b t a i n t h e t h i r d r o w o f A, t h a t i s$ $u_{3} = (\frac{4}{\sqrt{18}}, - \frac{1}{\sqrt{18}}, - \frac{1}{\sqrt{18}})$ $T h u s A = (\begin{matrix} \frac{1}{3} \frac{2}{3} \frac{2}{3} \\ 0 \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \\ \frac{4}{\sqrt{18}} - \frac{1}{\sqrt{18}} - \frac{1}{\sqrt{18}} \end{matrix})$ $W e e m p h a s i z e t h a t t h e a b o v e m a t r i x A$ $i s n o t u n i q u e .$
	Commented by bemath last updated on 06/Oct/20

	$waw . . . thanks a lot mr farmer math$
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