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Answer the following question: Without expanding determinant show that |0ab-a0c-b-c0| = 0 - Mathematics and Statistics

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Question

Answer the following question:

Without expanding determinant 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)|`

By taking ( –1) common from each of R1, R2, R3, we get,

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

By 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 4: Determinants and Matrices - Miscellaneous Exercise 4(A) [Page 77]

APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
Chapter 4 Determinants and Matrices
Miscellaneous Exercise 4(A) | Q II. (7) (iv) | Page 77

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