Question-57186

Question Number 57186 by Tawa1 last updated on 31/Mar/19

Commented by kaivan.ahmadi last updated on 31/Mar/19

| \begin{matrix} 1 1 1 \\ a b c \\ a^{2} b^{2} c^{2} \end{matrix} | \underset{- a^{2} R_{1} + R_{3}}{\overset{- {a R}_{1} + R_{2}}{\to}} a n d S a r r u s R u l e

| \begin{matrix} 1 1 1 \\ 0 b - a c - a \\ 0 b^{2} - a^{2} c^{2} - a^{2} \end{matrix} | | \begin{matrix} 1 1 1 \\ 0 b - a c - a \\ 0 b^{2} - a^{2} c^{2} - a^{2} \end{matrix} | =

[(b - a) (c^{2} - a^{2}) + 0 + 0] - [0 + 0 + (b^{2} - a^{2}) (c - a)] =

(b - a) (c - a) (c + a) - (b - a) (b + a) (c - a) =

(b - a) (c - a) [(c + a) - (b + a)] =

(b - a) (c - a) (c - b)

Commented by kaivan.ahmadi last updated on 31/Mar/19

t h i s d e t e r m i n a n t i s v a n d e r m l n d .

e a s l y i c h a n g e a = a_{1}, b = a_{2} a n d c = a_{3}

Commented by Tawa1 last updated on 31/Mar/19

God bless you sir

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Mar/19

∣ 0 0 1 ∣

∣ a_{1} - a_{3} a_{2} - a_{3} a_{3} ∣

∣ (a_{1} + a_{3}) (a_{1} - a_{3}) (a_{2} + a_{3}) (a_{2} - a_{3}) a_{3}^{2} ∣

(a_{1} - a_{3}) (a_{2} - a_{3}) ∣ 0 0 1 ∣

∣ 1 1 a_{3} ∣

∣ a_{1} + a_{3} a_{2} + a_{3} a_{3}^{2} ∣

(a_{1} - a_{3}) (a_{2} - a_{3}) (a_{2} + a_{3} - a_{1} - a_{3})

= (a_{2} - a_{1}) (a_{3} - a_{1}) (a_{3} - a_{2})

p r i n t i n g e r r o r i n p r o b l e m e l e m e n t

p r i n t e d {(a_{1}^{2})}_{3 \times 3} b u t s h o u l d b e {(a_{3}^{2})}_{3 \times 3}

Commented by Tawa1 last updated on 31/Mar/19

God bless you sir

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