Left| {,begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}{ca}&{ca' + c&#

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3 and 4 .Determinants and Matrices

hard

$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ =

A

$(ab - a'b')(bc - b'c')(ca - c'a')$

B

$(ab + a'b')(bc + b'c')(ca + c'a')$

C

$(ab' - a'b)(bc' - b'c)(ca' - c'a)$

D

$(ab' + a'b)(bc' + b'c)(ca' + c'a)$

Solution

ट्रिक : $a = 1,\,b = – 1,\,c = 0$, $a' = 2,\,b' = 2,\,c' = 1$

रखने पर, $\left| {\,\begin{array}{*{20}{c}}0&{ – 1}&2\\0&1&2\\{ – 1}&0&4\end{array}\,} \right| = 4$. विकल्प $(c) $ भी यही मान देता है।

Standard 12

Mathematics

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