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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
If $\left| {\,\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}\,} \right| = K{a^2}{b^2}{c^2},$ then $K = $
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Answer

c
(c) $\left| {\,\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}} \right| = abc\left| {\,\begin{array}{*{20}{c}}{ - a}&b&c\\a&{ - b}&c\\a&b&{ - c}\end{array}\,} \right|$

$ = (abc)(abc)\left| {\,\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right| = {a^2}{b^2}{c^2}( - 1)( - 4)$

$ = 4{a^2}{b^2}{c^2} = K{a^2}{b^2}{c^2}$,

$(given) ==> K = 4.$

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