Prove that 1 1 1 1 a b c d a2 b2 c2 d2 a3 b3 c3 d3 - a - ba - ca - db - cb - dc - d-

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Byju's Answer

Standard XII

Mathematics

Determinant

Prove that 1...

Question

Prove that
$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 & 1 a & b & c & d a^{2} & b^{2} & c^{2} & d^{2} a^{3} & b^{3} & c^{3} & d^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = (a - b) (a - c) (a - d) (b - c) (b - d) (c - d)$ .

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Solution

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

$= ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 & 0 a & b - a & c - a & d - a a^{2} & b^{2} - a^{2} & c^{2} - a^{2} & d^{2} - a^{2} a^{3} & b^{3} - a^{3} & c^{3} - a^{3} & d^{3} - a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ [C_{2} \to C_{2} - C_{1}; C_{3} \to C_{3} - C_{1}; C_{4} \to C_{4} - C_{1}]$

$= (b - a) (c - a) (d - a) ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 & 0 a & 1 & 1 & 1 a^{2} & a + b & a + c & a + d a^{3} & a^{2} + a b + b^{2} & a^{2} + a c + c^{2} & a^{2} + a d + d^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

$= (b - a) (c - a) (d - a) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a + b & a + c & a + d a^{2} + a b + b^{2} & a^{2} + a c + c^{2} & a^{2} + a d + d^{2} \end{matrix} ∣ ∣ ∣ ∣$

$= (b - a) (c - a) (d - a) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 a + b & c - b & d - b a^{2} + a b + b^{2} & (c - b) (a + b + c) & (d - b) (a + b + d) \end{matrix} ∣ ∣ ∣ ∣ ∣ [C_{2} \to C_{2} - C_{1}; C_{3} \to C_{3} - C_{1}]$

$= (b - a) (c - a) (d - a) (c - b) (d - b) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 a + b & 1 & 1 a^{2} + a b + b^{2} & a + b + c & a + b + d \end{matrix} ∣ ∣ ∣ ∣$
$= (b - a) (c - a) (d - a) (c - b) (d - b) (a + b + d - a - b - c) = (a - b) (a - c) (a - d) (b - c) (b - d) (c - d)$

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Similar questions

D is a point on the side BC of $Δ A B C$ such that $∠ A D C = ∠ B A C$ . Prove that $\frac{C B}{C A}$ = $\frac{C A}{D C}$ or $C A^{2}$ = $C B \times C D$ .

[2 MARKS]

Q. Prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + a b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3. (b - c) (c - a) (a - b) (a + b + c) (a b + b c + c a)

Q. Prove the following identities:

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & b c + a d & b^{2} c^{2} + a^{2} d^{2} 1 & c a + b d & c^{2} a^{2} + b^{2} d^{2} 1 & a b + c d & a^{2} b^{2} + c^{2} d^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - (b - c) (c - a) (a - b) (a - d) (b - d) (c - d)

Q. Prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} b + c - a - d & b c - a d & b c (a + d) - a d (b + c) c + a - b - d & c a - b d & c a (b + d) - b d (c + a) a + b - c - d & a b - c d & a b (c + d) - c d (a + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 2 (b - c) (c - a) (a - b) (a - d) (b - d) (c - d)

Q. Show that :

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + c b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 (b - c) (c - a) (a - b) (a + b + c) (b c + c a + a b)

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Prove that∣∣ ∣ ∣ ∣∣1111abcda2b2c2d2a3b3c3d3∣∣ ∣ ∣ ∣∣=(a−b)(a−c)(a−d)(b−c)(b−d)(c−d).

Prove that
$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 & 1 a & b & c & d a^{2} & b^{2} & c^{2} & d^{2} a^{3} & b^{3} & c^{3} & d^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = (a - b) (a - c) (a - d) (b - c) (b - d) (c - d)$ .