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If `b^(2) -ac lt 0 and a gt 0` then the value of the determinant isA. positiveB. negativeC. zeroD. `b^(2)+ae` |
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Answer» Correct Answer - 2 We have `|{:(a,b,ax+by),(b,c,bx+cy),(ax+by,bx+cy,0):}|` `|{:(" "a," "b," "0),(" "b," "c," "0),(ax+by,bx+cy,-(ax^(2)+2bxy+cy^(2))):}|` [Applying `C_(3)toC_(3)-xC_(1)-yC_(2)`] `=-(ax^(2)+2bx+cy^(2))(ac-b^(2))` `(1)/(a)(b^(2)-ac)[(ax+by)^(2)+Y^(2)(ac-b^(2))]lt0` `[therefore b^(2)-aclt 0 and agt 0]` |
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