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The value of determinant abbcabacabcaabc|a-bb+cab-ac+abc-aa+bc| is ______. - Mathematics

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Question

The value of determinant `|("a" - "b", "b" + "c", "a"),("b" - "a", "c" + "a", "b"),("c" - "a", "a" + "b", "c")|` is ______.

Options

  • a3 + b3 + c3

  • 3bc

  • a3 + b3 + c3 – 3abc

  • None of these

MCQ
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Solution

The value of determinant `|("a" - "b", "b" + "c", "a"),("b" - "a", "c" + "a", "b"),("c" - "a", "a" + "b", "c")|` is none of these.

Explanation:

Here, we have `|("a" - "b", "b" + "c", "a"),("b" - "a", "c" + "a", "b"),("c" - "a", "a" + "b", "c")|`

C2 → C2 + C3

⇒ `|("a" - "b", "a" + "b" + "c", "a"),("b" - "a", "a" + "b" + "c", "b"),("c" - "a", "a" + "b" + "c", "c")|`

⇒ `("a" + "b" + "c") |("a" - "b", 1, "a"),("b" - "a", 1, "b"),("c" - "a", 1, "c")|`  .....(Taking a + b + c common from C2)

R1 → R1 – R2, R2 → R2 – R3

⇒ `("a" + "b" + "c") |(2("a" - "b"), 0, "a" - "b"),("b" - "c", 0, "b" - "c"),("c" - "a", 1, "c")|`

Taking (a – b) and (b – c) common from R1 and R2 respectively

⇒ `("a" + "b" + "c")("a" - "b")("b" - "c") |(2, 0, 1),(1, 0, 1),("c" - "a", 1, "c")|`

Expanding along C2

⇒ `("a" + "b" + "c")("a" - "b")("b" - "c") [-1|(2, 1),(1, 1)|]`

⇒ (a + b + c)(a – b)(b – c)(– 1)

⇒ (a + b + c)(a – b)(c – b)

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Chapter 4: Determinants - Exercise [Page 80]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 4 Determinants
Exercise | Q 25 | Page 80

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