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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}}{b + c}&{a - b}&a\\{c + a}&{b - c}&b\\{a + b}&{c - a}&c\end{array}\,} \right| = $
  • A
    ${a^3} + {b^3} + {c^3} - 3abc$
  • ✓
    $3abc - {a^3} - {b^3} - {c^3}$
  • C
    ${a^3} + {b^3} + {c^3} - {a^2}b - {b^2}c - {c^2}a$
  • D
    $(a+b+c)(a^2+b^2+c^2+ab+bc+ca)$
✓

Answer

Correct option: B.
$3abc - {a^3} - {b^3} - {c^3}$
b
(b) $\Delta = \left| {\,\begin{array}{*{20}{c}}{2(a + b + c)}&0&{a + b + c}\\{c + a}&{b - c}&b\\{a + b}&{c - a}&c\end{array}\,} \right|$

by ${R_1} \to {R_1} + {R_2} + {R_3}$

$\Delta = (a + b + c)\,.\,\left| {\,\begin{array}{*{20}{c}}2&0&1\\{c + a}&{b - c}&b\\{a + b}&{c - a}&c\end{array}\,} \right|$

On expanding,

$ - (a + b + c)\,({a^2} + {b^2} + {c^2} - ab - bc - ca)$

= $ -(a^3  + b^3 + c^3  -  3abc) = 3abc - a^3 -b^3 - c^3$

Trick : Put $a = 1,\,b = 2,\,c = 3$ and check it.

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