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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
The value of the determinant $\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right|$ is equal to 
  • A
    ${a^3} + {b^3} + {c^3}$
  • B
    ${a^3} - {b^3} - {c^3}$
  • ✓
    $0$
  • D
    $ - {a^3} + {b^3} + {c^3}$
✓

Answer

Correct option: C.
$0$
c
(c) $\left| {\,\begin{array}{*{20}{c}}0&{{b^3} - {a^3}}&{{c^3} - {a^3}}\\{{a^3} - {b^3}}&0&{{c^3} - {b^3}}\\{{a^3} - {c^3}}&{{b^3} - {c^3}}&0\end{array}\,} \right|$

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

$[{C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}]$ and then taking out common

$({b^2} - {a^3})$ from $II^{nd}$ column and ( ${c^3} - {a^3}$) from $III^{rd}$ column].

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