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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
$\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
    ${a^3} + {b^3} + {c^3} - 3abc$
  • ✓
    None of these
✓

Answer

Correct option: D.
None of these
d
(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)$.

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