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Using properties of determinants, prove ...

Using properties of determinants, prove the following : `|a a^2b c bb^2c a cc^2a b|=(a-b)(b-c)(c-a)(b c+c a+a b)`

Text Solution

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Let the value of the given determinant be `Delta`. Then,
`Delta = |{:(a, b, c),(a^(2), b^(2), c^(2)),(bc, ca, ab):}|`
`= |{:(a-c, b-c, c),(a^(2)-c^(2), b^(2)-c^(2), c^(2)),(b(c-a), a(c-b), ab):}|" "[C_(1) to (C_(1) -C_(3)), C_(2) to (C_(2)-C_(3))]`
`=(a-c)(b-c)* |{:(1, 1, c),(a+c, b+c, c^(2)),(-b, -a, ab):}|`
`["taking (a-c) common from"C_(1) " and (b-c) common from"C_(2)]`
`=(a-c)(b-c)* |{:(1, 0, 0),(a+c, b-a, -ca),(-b, b-a, (a+c)b):}| [C_(2) to (C_(2) -C_(1)), C_(3) to (C_(3)-cC_(1))]`
`=(a-c)(b-c)* 1*|{:(b-a, -ca),(b-a, (a+c)b):}|{"expanded by"R_(1)}`
` =(a-c)(b-c)*(b-c)|{:(1, -ca), (1, (a+c)b):}|`
` = (a-b)(b-c)(c-a)(ab+bc+ca)`
Hence, `Delta = (a-b) (b-c)(c-a)(ab+bc+ca)`
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