If determinant (((a a^2 1+a^3 )),((b b^2 1+b^3 )),((c c^2 1+c^3 )))= 0 a≠b≠c → { ((a =?)),((b=? )),((c=?)) :}


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Question Number 115508 by bemath last updated on 26/Sep/20
$If determinant (((a a^2 1+a^3 )),((b b^2 1+b^3 )),((c c^2 1+c^3 )))= 0 a≠b≠c → { ((a =?)),((b=? )),((c=?)) :}$
$I f \| \begin{matrix} a a^{2} 1 + a^{3} \\ b b^{2} 1 + b^{3} \\ c c^{2} 1 + c^{3} \end{matrix} \| = 0$ $a \neq b \neq c \to {\begin{cases} a = ? \\ b = ? \\ c = ? \end{cases}$
	Answered by bobhans last updated on 26/Sep/20

	$\Rightarrow \| \begin{matrix} a a^{2} 1 \\ b b^{2} 1 \\ c c^{2} 1 \end{matrix} \| + \| \begin{matrix} a a^{2} a^{3} \\ b b^{2} b^{3} \\ c c^{2} c^{3} \end{matrix} \| = 0$ $\Rightarrow (- 1) \| \begin{matrix} 1 a^{2} a \\ 1 b^{2} b \\ 1 c^{2} c \end{matrix} \| + a b c \| \begin{matrix} 1 a a^{2} \\ 1 b b^{2} \\ 1 c c^{2} \end{matrix} \| = 0$ $\Rightarrow {(- 1)}^{2} \| \begin{matrix} 1 a a^{2} \\ 1 b b^{2} \\ 1 c c^{2} \end{matrix} \| + a b c \| \begin{matrix} 1 a a^{2} \\ 1 b b^{2} \\ 1 c c^{2} \end{matrix} \| = 0$ $\Rightarrow (1 + a b c) \| \begin{matrix} 1 a a^{2} \\ 1 b b^{2} \\ 1 c c^{2} \end{matrix} \| = 0$ $\Rightarrow (1 + a b c) \| \begin{matrix} 1 a a^{2} \\ 0 b - a b^{2} - a^{2} \\ 0 c - a c^{2} - a^{2} \end{matrix} \| = 0$ $\Rightarrow (1 + a b c) \| \begin{matrix} b - a b^{2} - a^{2} \\ c - a c^{2} - a^{2} \end{matrix} \| = 0$ $\Rightarrow (1 + a b c) [(b - a) (c - a) (c + a) - (c - a) (b - a) (b + a)] = 0$ $\Rightarrow (1 + a b c) (b - a) (c - a) [c + a - b - a] = 0$ $(1 + a b c) (b - a) (c - a) (c - b) = 0$ $s i n c e a \neq b \neq c; s o w e g e t a b c = - 1$ $l e t a = k, b = l \to c = - \frac{1}{k . l}$
	Commented by bemath last updated on 26/Sep/20

	$g a v e k u d o s$
	Answered by TANMAY PANACEA last updated on 26/Sep/20

	$\| \begin{matrix} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{matrix} \| + \| \begin{matrix} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{matrix} \| = 0$ $\| \begin{matrix} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{matrix} \| + a b c \| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \| = 0$ $\| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \| + a b c \| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \| = 0$ $D + a b c \times D = 0$ $D (1 + a b c) = 0$ $D \neq 0 s o a b c = - 1$ $l e t a = k b = \frac{1}{k} c = - 1$ $i h a v e j u s t t r i e d . . .$
	Commented by bemath last updated on 26/Sep/20

	$g a v e k u d o s s i r .$
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