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 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 \dots

Answered by math1967 last updated on 02/Oct/18

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 \dots

Commented by rahul 19 last updated on 02/Oct/18

thanks sir ��.