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Without expanding the determinants, show that |0ab-a0c-b-c0| = 0

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Question

Without expanding the determinants, show that `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0

Sum
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Solution

Let D = `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|`

Taking (– 1) common from R1, R2, R3, we get

D = `(- 1)^3 |(0, -"a", -"b"),("a", 0, -"c"),("b", "c", 0)|`

Interchanging rows and columns, we get

D = `-1|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|`

∴ D = – 1(D)
∴ 2D = 0
∴ D = 0

∴ `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0

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Chapter 6: Determinants - MISCELLANEOUS EXERCISE - 6 [Page 95]

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Balbharati Mathematics and Statistics (Commerce) Part 1 [English] Standard 11 Maharashtra State Board
Chapter 6 Determinants
MISCELLANEOUS EXERCISE - 6 | Q 4) iv) | Page 95

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