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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
If $a,b,c$ are different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$, then
  • A
    $a + b + c = 0$
  • ✓
    $abc = 1$
  • C
    $a + b + c = 1$
  • D
    $ab + bc + ca = 0$
✓

Answer

Correct option: B.
$abc = 1$
b
(b) $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$==> $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3}}\\b&{{b^2}}&{{b^3}}\\c&{{c^2}}&{{c^3}}\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\b&{{b^2}}&1\\c&{{c^2}}&1\end{array}\,} \right| = 0$
==> $abc\,\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| - \left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = 0$
==> $(abc - 1)\,\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = 0$
Since $a,b,c$ are different, so $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| \ne 0$
Hence $abc - 1 = 0$ i.e., $abc = 1$.

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