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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
$\left| {\,\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}\,} \right| = $
  • A
    $3abc + {a^3} + {b^3} + {c^3}$
  • ✓
    $3abc - {a^3} - {b^3} - {c^3}$
  • C
    $abc - {a^3} + {b^3} + {c^3}$
  • D
    $abc + {a^3} - {b^3} - {c^3}$
✓

Answer

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

                                                       $({R_1} \to {R_1} + {R_2} + {R_3})$

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

= $3abc - {a^3} - {b^3} - {c^3}$,                    (After simplification).

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