3 and 4 .Determinants and Matrices
easy

$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $

A

$0$

B

$(a - b)(b - c)(c - a)$

C

${a^3} + {b^3} + {c^3} - 3abc$

D

None of these

Solution

(d) $\left| {\,\begin{array}{*{20}{c}}{a – 1}&a&{bc}\\{b – 1}&b&{ca}\\{c – 1}&c&{ab}\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}a&a&{bc}\\b&b&{ca}\\c&c&{ab}\end{array}\,} \right| – \left| {\,\begin{array}{*{20}{c}}1&a&{bc}\\1&b&{ca}\\1&c&{ab}\end{array}\,} \right|$

= $ – \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\b&{{b^2}}&1\\c&{{c^2}}&1\end{array}\,} \right| = – \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\{b – a}&{{b^2} – {a^2}}&0\\{c – a}&{{c^2} – {a^2}}&0\end{array}\,} \right|$

                                               [By ${R_2} \to {R_2} – {R_1};\,{R_3} \to {R_3} – {R_1}$]

= $ – (a – b)\,(b – c)\,(c – a)$.

Standard 12
Mathematics

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