MCQ
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 = $
- A$-4$
- B$2$
- ✓$4$
- D$8$
✓
Answer
Correct option: C.
$4$
c
(c) $\left| {\,\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}} \right| = abc\left| {\,\begin{array}{*{20}{c}}{ - a}&b&c\\a&{ - b}&c\\a&b&{ - c}\end{array}\,} \right|$
(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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