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Question Number 57306 by Aditya789 last updated on 02/Apr/19

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Apr/19

$i s i t a b o r - a b$
Commented by Aditya789 last updated on 02/Apr/19

$ab$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Apr/19

	$△ = d e t e r m i n a n t$ $t h e d e g r e e o f △ = 6$ $n o w w h e n w e p u t a = 0 △ = 0 s o a i s f a c t o r o f △$ $s i m i l a r l y i f b = 0 t h e n △ = 0 s o b i s f a c t o r o f △$ $w h e n c = 0 t h e n △ = 0 s o c i s f a c t o r o f △$ $n o w$ $p u t a + b + c = 0 t h a t m e a n s i n p l a c e o f (b + c) p u t (- a)$ $i n p l a c e o f (c + a) p u t (- b) a n d i n p l a c e o f (a + b) p u t$ $(- c)$ $△ =∣ {(b + c)}^{2} b a c a ∣ (p u t a + b + c = 0)$ $∣ a b {(c + a)}^{2} b c ∣$ $∣ a c b c {(a + b)}^{2} ∣$ $= ∣ {(- a)}^{2} b a c a ∣$ $∣ a b {(- b)}^{2} b c ∣$ $∣ a c b c {(- c)}^{2} ∣$ $a b c ∣ a b c ∣$ $∣ a b c ∣$ $∣ a b c ∣$ $l o o k t h r e e r o w s$ $s o △ = 0$ $s o (a + b + c) i s a l s o a f a c t o r . . .$ $n o w w h e n w e p u t (a + b + c) = 0$ $t h r e e r o w s b e c o m d i d e n t i c a l$ $h e n c e b y r u l e {(a + b + c)}^{2} i s a l s o a f a c t o r$ $△ = k a b c (a + b + c) {(a + b + c)}^{2}$ $△ = k a b c {(a + b + c)}^{3}$ $n o w o u r t a s k i s t o f i n d v a l u e o f k$ $△ =∣ {(b + c)}^{2} b a c a ∣$ $∣ a b {(c + a)}^{2} b c ∣= k a b c {(a + b + c)}^{3}$ $∣ a c b c {(a + b)}^{2} ∣$ $p u t a = 1, b = 1 a n d c = 1 b o t h s i d e$ $∣ 4 1 1 ∣$ $∣ 1 4 1 ∣$ $∣ 1 1 4 ∣ = k \times 1 \times 1 \times 1 {(1 + 1 + 1)}^{3}$ $54 = 27 k s o k = 2$ $s o a n s w e r o f d e t e r m i n a t i s 2 a b c {(a + b + c)}^{3}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Apr/19

	Commented by rahul 19 last updated on 02/Apr/19

	$V e r y h e l p f u l s i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Apr/19

	$t h a n k y o u . . . s o u r c e h i g h e r a l g e b r a$ $1) B e r n a r d a n d c h i l d$ $2) H a l l a n d k n i g h t$ $f a m o u s b o o k s . . .$
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