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-

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