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Using properties of determinants, prove that `|{:(0, ab^(2), ac^(2)),(a^(2)b, 0, bc^(2)),(a^(2)c, cb^(2), 0):}|=2a^(3)b^(3)c^(3)`

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


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