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Using properties of determinant show that |a+babaa+ccbcb+c| = 4abc - Mathematics and Statistics

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प्रश्न

Using properties of determinant show that

`|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc

बेरीज
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उत्तर

L.H.S. = `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|`

By C1 + (C2 + C3), we get,

L.H.S. = `|(2("a" + "b"), "a", "b"),(2("a" + "c"), "a" + "c", "c"),(2("b" + "c"), "c", "b" + "c")|`

By taking 2 common from C1, we get,

L.H.S. = `2|("a" + "b", "a", "b"),("a" + "c", "a" + "c", "c"),("b" + "c", "c", "b" + "c")|`

By C1 – (C2 + C3), we get,

L.H.S. = `2|(0, "a", "b"),(-"c", "a" + "c", "c"),(-"c", "c", "b" + "c")|`

By C2 + C1 and C3 + C1, we get

L.H.S. = `2|(0, "a", "b"),(-"c", "a", 0),(-"c", 0, "b")|`

By taking c, a, b common from C1, C2, C3 respectively, we get,

L.H.S. = `2("c"  "a"  "b")|(0, 1, 1),(-1, 1, 0),(-1, 0, 1)|`

= 2abc [0 (1 – 0) – 1 ( – 1 + 0) + 1 (0 + 1)]

= 2abc (0 + 1 + 1)

= 4abc

= RHS.

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पाठ 4: Determinants and Matrices - Exercise 4.2 [पृष्ठ ६८]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 4 Determinants and Matrices
Exercise 4.2 | Q 3. (i) | पृष्ठ ६८

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