Determine the value of the following proposition . ( True or False ) ∃ x ∈ R ; determinant ((( 1+2x),( 2x),(2x)),(( 2x),( 1+2x),( 2x )),(( 2x),( 2x),(1 +2x)))= x^( 3) + 8x−2 −−−−−−−−−


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Question Number 161369 by mnjuly1970 last updated on 17/Dec/21

$D e t e r m i n e t h e v a l u e o f t h e f o l l o w i n g$ $p r o p o s i t i o n . (T r u e o r F a l s e)$ $\exists x \in R; \| \begin{matrix} 1 + 2 x & 2 x & 2 x \\ 2 x & 1 + 2 x & 2 x \\ 2 x & 2 x & 1 + 2 x \end{matrix} \| = x^{3} + 8 x - 2$ $- - - - - - - - -$
	Answered by 1549442205PVT last updated on 17/Dec/21

	$△ = {(1 + 2 x)}^{3} + 4 x^{2} (1 + 2 x) + 8 x^{3} - 4 x^{2} (1 + 2 x) - 4 x^{2} (1 + 2 x) - 2 x {(1 + 2 x)}^{2}$ $= 1 + 6 x + 12 x^{2} + 8 x^{3} + 8 x^{3} - 4 x^{2} (1 + 2 x)$ $- 2 x - 8 x^{2} - 8 x^{3} = 4 x + 1$ $H e n c e, a b o v e r e s u l t i s f a l s e$ $w e c a n u s i n g e q u i v a l e n t t r a n s f o r m o f$ $t w o d e t e r m i n a n t f o l l o w a s :$ $\| \begin{matrix} 1 + 2 x 2 x 2 x \\ 2 x 1 + 2 x 2 x \\ 2 x 1 + 2 x 1 + 2 x \end{matrix} \| \sim \| \begin{matrix} 1 + 2 x 2 x 2 x \\ 2 x 1 + 2 x 2 x \\ 0 0 1 \end{matrix} \|$ $\sim \| \begin{matrix} 1 + 2 x 2 x 2 x \\ - 1 1 0 \\ 0 0 1 \end{matrix} \| . N o w w e c a l c u l a t e t h e v a l u e o f l a s t d e t e r m i n a n t :$ $△ = (1 + 2 x) . 1.1 + 2 x . 0.0 + 0 . (- 1) . 2 x$ $- 0.1.2 x - (- 1) . 2 x . 1 - 0.0 . (1 + 2 x)$ $= 2 x + 1 + 2 x = 4 x + 1$
	Answered by MJS_new last updated on 17/Dec/21

	$\Rightarrow 4 x + 1 = x^{3} + 8 x - 2$ $\Leftrightarrow x^{3} + 4 x - 3 = 0$ $x = \sqrt[3]{\frac{3}{2} + \frac{\sqrt{1497}}{18}} - \sqrt[3]{- \frac{3}{2} + \frac{\sqrt{1497}}{28}} \approx . 673593058$ $\Rightarrow true because indeed \exists x \in R$
	Commented by 1549442205PVT last updated on 18/Dec/21

	$T h a n k y o u S i r M j s N e w, i {d i d n}^{'} t u n d e r s t a n d t h e q u e s t i o n$ $a n d g i v e o u t a w r o n g c o n c l u s i o n$
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