Prove that One factor of determinant (((a^2 +x),( ab),( ac)),(( ab),(b^2 +x),( cb)),(( ca),( cb),(c^2 +x))) is x^2 .


Question and Answers Forum

All Questions Topic List
UNKNOWN Questions
Previous in All Question Next in All Question
Previous in UNKNOWN Next in UNKNOWN
Question Number 44612 by rahul 19 last updated on 02/Oct/18

$P r o v e t h a t One factor of \| \begin{matrix} a^{2} + x & a b & a c \\ a b & b^{2} + x & c b \\ c a & c b & c^{2} + x \end{matrix} \| is x^{2} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

	Commented by $@ty@m last updated on 02/Oct/18

	$i t i s c o r r e c t .$ $b u t i t i s e x p e c t e d t o u s e p r o p e r t i e s$ $o f d e t e r m i n a n t t o s o l v e p r o b l e m s$ $o f d e t e r m i n a n t c h a p t e r .$
	Commented by rahul 19 last updated on 02/Oct/18

	$Y e s s i r, i w a n t t h i s Q . t o b e s o l v e d b y$ $p r o p e r t i e s .$ $(A l t h o u g h t h a n k s f o r t h i s a l s o) .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

	$l e t △ = x^{2} f (a, b, c)$ $n o w u s i n g f a c t o r t h e o r e m$ $p (x) = (x - α) q (x)$ $p (α) = 0 t h a t m e a n s x - α i s s f a c t o r o f p (x)$ $u s i n g s a m e p r i n c i p l e w e p u t x = 0 i n △$ $n x t t a k e c o m m o n f r o m 1 s t r o w$ $b f r o m s e c o n d r o w$ $c f r o m t h i r d r o w . .$ $i t i s c l e a r t h a t t h r e e r o w i d e n t i c a l$ $s o △ = 0$ $h e n c e p r o v e d . . .$
	Answered by math1967 last updated on 02/Oct/18
	$abc determinant (((a+(x/a)),b,c),(a,(b+(x/b)),c),(a,b,(c+(x/(c )) ))) =abc determinant (((x/a),(−(x/b) ),0),(0,(x/b),(−(x/c))),(a,b,(c+(x/c))))R_1 ′→R_1 −R_2 ,R_(2 ) ′→R_(2 ) −R_3 abcx^2 determinant (((1/a),(−(1/b)),0),(0,(1/b),(−(1/c))),(a,b,(c+(x/c)))) ((abcx^2 )/(ab)) determinant ((1,(−1),0),(0,1,(−(1/c))),(a^2 ,b^2 ,(c+(x/c)))) cx^2 determinant ((1,0,0),(0,1,(−(1/c))),(a^2 ,(a^2 +b^2 ),(c+(x/c))))C_2 ′→C_1 +C_(2 ) cx^2 {1×(c+(x/c))+((a^2 +b^2 )/c)} =x^2 (x+a^2 +b^2 +c^2 ) ∴x^2 is factor...$
	$a b c \| \begin{matrix} a + \frac{x}{a} & b & c \\ a & b + \frac{x}{b} & c \\ a & b & c + \frac{x}{c} \end{matrix} \|$ $= a b c \| \begin{matrix} \frac{x}{a} & - \frac{x}{b} & 0 \\ 0 & \frac{x}{b} & - \frac{x}{c} \\ a & b & c + \frac{x}{c} \end{matrix} \| R_{1}^{'} \to R_{1} - R_{2}, R_{2}^{'} \to R_{2} - R_{3}$ ${a b c x}^{2} \| \begin{matrix} \frac{1}{a} & - \frac{1}{b} & 0 \\ 0 & \frac{1}{b} & - \frac{1}{c} \\ a & b & c + \frac{x}{c} \end{matrix} \|$ $\frac{{a b c x}^{2}}{a b} \| \begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - \frac{1}{c} \\ a^{2} & b^{2} & c + \frac{x}{c} \end{matrix} \|$ ${c x}^{2} \| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{c} \\ a^{2} & a^{2} + b^{2} & c + \frac{x}{c} \end{matrix} \| C_{2}^{'} \to C_{1} + C_{2}$ ${c x}^{2} {1 \times (c + \frac{x}{c}) + \frac{a^{2} + b^{2}}{c}}$ $= x^{2} (x + a^{2} + b^{2} + c^{2})$ $∴ x^{2} i s f a c t o r . . .$
	Commented by rahul 19 last updated on 02/Oct/18
	thanks sir ��.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum