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If xkxkxkykykykzkzkzk|xkxk+2xk+3ykyk+2yk+3zkzk+2zk+3| = (x - y) (y - z) (z - x)xyz(1x+1y+1z) then

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Question

If `|("x"^"k", "x"^("k" + 2), "x"^("k" + 3)),("y"^"k", "y"^("k" + 2), "y"^("k" + 3)),("z"^"k", "z"^("k" + 2), "z"^("k" + 3))|` = (x - y) (y - z) (z - x)`(1/"x"+ 1/"y" + 1/"z") ` then

Options

  • k = –3

  • k = –1

  • k = 1

  • k = 3

MCQ
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Solution

k = – 1

Explanation:

L.H.S →

`|(x^k, x^(k + 2), x^(k + 3)),(y^k, y^(k + 2), y^(k + 3)),(z^k, z^(k + 2), z^(k + 3))| = x^k y^k z^k |(1, x^2, x^3),(1, y^2, y^3),(1, z^2, z^3)|`

= (xyz)k(x – y) (y – z) (z – x) (xy + yz + zx)

R.H.S. →

= `(x - y) (y - z) (z - x) (1/x+ 1/y + 1/z)`

= `(x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`

L.H.S. = R.H.S.

∴ `(xyz)^k (x - y) (y - z) (z - x) (xy + yz + zx) = (x - y) (y - z) (z - x) ((xy + yz + zx)/(xyz))`

∴ `(xyz)^k cancel((x - y)) cancel((y - z)) cancel((z - x)) (xy + yz + zx) = cancel((x - y)) cancel((y - z)) cancel((z - x)) ((xy + yz + zx)/(xyz))`

∴ `(xyz)^k  cancel((xy + yz + zx)) = cancel((xy + yz + zx))/(xyz)`

∴ (xyz)k = `1/"xyz"`

∴ (xyz)k = xyz- 1

∴ k = – 1

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Chapter 4: Determinants and Matrices - Miscellaneous Exercise 4(A) [Page 75]

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Balbharati Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
Chapter 4 Determinants and Matrices
Miscellaneous Exercise 4(A) | Q I. (2) | Page 75

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