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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^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $
  • ✓
    $4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
  • B
    $3\,\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
  • C
    $2\,\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
  • D
    None of these
✓

Answer

Correct option: A.
$4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$
a
(a) Apply ${R_2} - {R_3}$ and note that ${(x + y)^2} - {(x - y)^2} = 4xy$

$\therefore $ $\Delta = 4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right|$

= $4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$

{Applying ${R_3} - ({R_1} - 2{R_2}) $}

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