Prove that- - 1 1 1- a b c- a3 b3 c3 - - a-bb-cc-aa-b-c

LHS: [ 1 amp; 1 amp;1; a amp; b amp; c; a^3 amp; b^3 amp; c^3 ] =1·[ b amp;c; b^3 amp;c^3 ] 1·[ a amp;c ...

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Standard XII

Mathematics

Summation by Sigma Method

Prove that| ...

Question

Prove that
| $⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ⎤ ⎥ ⎦$ | = $(a - b) (b - c) (c - a) (a + b + c)$

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Solution

LHS:

$⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ⎤ ⎥ ⎦$

$= 1 \cdot det (\begin{matrix} b & c b^{3} & c^{3} \end{matrix}) - 1 \cdot det (\begin{matrix} a & c a^{3} & c^{3} \end{matrix}) + 1 \cdot det (\begin{matrix} a & b a^{3} & b^{3} \end{matrix})$

$= 1 \cdot (b c^{3} - c b^{3}) - 1 \cdot (a c^{3} - c a^{3}) + 1 \cdot (a b^{3} - b a^{3})$

$= b c^{3} - b^{3} c - a c^{3} + a^{3} c + a b^{3} - a^{3} b$

RHS:

$(a - b) (b - c) (c - a) (a + b + c)$

$= (a b - a c - b^{2} + b c) (c - a) (a + b + c)$

$= (a b c - a^{b} - a c^{2} + a^{2} c - b^{2} c + a b^{2} + b c^{2} - a b c) (a + b + c)$

$= a^{2} b^{2} + a b^{3} + a b^{2} c - a^{3} b - a^{2} b^{2} - a^{2} b c + a^{3} c + a^{2} b c + a^{2} c^{2} - a^{2} c^{2} - a b c^{2} - a c^{3} + a b c^{2} + b^{2} c^{2} + b c^{3} - a b^{2} c - b^{3} c - b^{2} c^{2}$

$= b c^{3} - b^{3} c - a c^{3} + a^{3} c + a b^{3} - a^{3} b$

LHS=RHS

Hence proved.

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Similar questions

Q. Solve

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By using properties of determinants, show that:

(i)

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

Prove that : =

(a - b) (b - c) (c - a) (a + b + c) .

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$In any Δ A B C, prove that : \frac{b s e c B + c s e c C}{t a n B + t a n C} = \frac{c s e c C + a s e c A}{t a n C + t a n A} = \frac{a s e c A + b s e c B}{t a n A + t a n B}$

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Prove that|⎡⎢⎣111abca3b3c3⎤⎥⎦| = (a−b)(b−c)(c−a)(a+b+c)

Prove that
| $⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ⎤ ⎥ ⎦$ | = $(a - b) (b - c) (c - a) (a + b + c)$