- Home
- Standard 12
- Mathematics
$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $
$0$
$(a - b)(b - c)(c - a)$
${a^3} + {b^3} + {c^3} - 3abc$
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)$.