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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
If $a, b, c$  are all different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} - 1}\\b&{{b^3}}&{{b^4} - 1}\\c&{{c^3}}&{{c^4} - 1}\end{array}\,} \right|$ = $0$ , then the value of $abc(ab + bc + ca)$ is
  • ✓
    $a + b + c$
  • B
    $0$
  • C
    ${a^2} + {b^2} + {c^2}$
  • D
    ${a^2} - {b^2} + {c^2}$
✓

Answer

Correct option: A.
$a + b + c$
a
(a) $\left[ {\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4} - 1}\\b&{{b^3}}&{{b^4} - 1}\\c&{{c^3}}&{{c^4} - 1}\end{array}} \right]\, = 0$

or $\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4}}\\b&{{b^3}}&{{b^4}}\\c&{{c^3}}&{{c^4}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ - 1}\\b&{{b^3}}&{ - 1}\\c&{{c^3}}&{ - 1}\end{array}\,} \right| = 0$

or $abc{\rm{ }}\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\1&{{b^2}}&{{b^3}}\\1&{{c^2}}&{{c^3}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ - 1}\\{a - b}&{{a^3} - {b^3}}&0\\{a - c}&{{a^3} - {c^3}}&0\end{array}\,} \right|\, = 0$

or $abc{\rm{ }}\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\0&{{a^2} - {b^2}}&{{a^3} - {b^3}}\\0&{{a^2} - {c^2}}&{{a^3} - {c^3}}\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ - 1}\\{a - b}&{{a^3} - {b^3}}&0\\{(a - c)}&{({a^3} - {c^3})}&0\end{array}\,} \right|\, = 0$

or $(abc)\,(a - b)\,(a - c)\,\left| {\,\begin{array}{*{20}{c}}1&{{a^2}}&{{a^3}}\\0&{a + b}&{{a^2} + {b^2} + ab}\\0&{a + c}&{{a^2} + {c^2} + ac}\end{array}\,} \right|\, + $

$(a - b)\,(a - c)\,\left| {\,\begin{array}{*{20}{c}}a&{{a^3}}&{ - 1}\\1&{{a^2} + {b^2} + ab}&0\\1&{{a^2} + {c^2} + ac}&0\end{array}\,} \right|$

or $(a - b)\,(a - c)\,[(abc)[(a + b)\,({a^2} + {c^2} + ac) - $

$(a + c)({a^2} + {b^2} + ab)]] + ( - 1)\,(a - b)\,(a - c)$

$[{a^2} + {c^2} + ac - {a^2} - {b^2} - ab] = 0$

= $(abc)\,[(a - b)\,(a - c)\,(c - b)(ac + ab + bc)]$

$ + ( - 1)(a - b)(a - c)(c - b)(a + b + c) = 0$

$ \Rightarrow $ $(abc)\,(ac + ab + bc) = a + b + c$.

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