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the determinant `Delta=|[a^2+x, ab, ac] , [ab, b^2+x, bc] , [ac, bc, c^2+x]|` is divisible byA. xB. `x^(2)`C. `x^(3)`D. none of these

Answer» Correct Answer - A::B
`Delta =|{:(a^(3)+ax,,ab,,ac),(a^(2)b,,b^(2)+x,,bc),(a^(2)c,,bc,,c^(2)+x):}|`
Applying `C_(1) to C_(1) +bC_(2) +cC_(3)" and taking " a^(2)+b^(2)+c^(2)+x` common we get
`Delta =(1)/(a)(a^(2) +b^(2) +c^(2)+x ) |{:(a,,ab,,ac),(b,,b^(2)+x,,bc),(c,,bc,,c^(2)+x):}|`
Applying `C_(2) to C_(2) -bC_(1) " and " C_(3) to C_(3) -cC_(1) ` we get
`Delta =(1)/(a) (a^(2) +b^(2)+c^(2)+x) |{:(a,,0,,0),(b,,x,,0),(c,,0,,x):}|`
`=(1)/(a) (a^(2) +b^(2)+c^(2) +x) (ax^(2))`
`= x^(2) (a^(2) +b^(2)+c^(2) +x)`
Thus `Delta ` is divisible by x and `x^(2)`


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