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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}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right| = $
  • A
    ${a^2} + {b^2} + {c^2}$
  • B
    $(a + b)\,(b + c)\,(c + a)$
  • ✓
    $(a - b)(b - c)(c - a)$
  • D
    None of these
✓

Answer

Correct option: C.
$(a - b)(b - c)(c - a)$
c
(c) $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}\,} \right|\, = \left| {\,\begin{array}{*{20}{c}}0&{a - b}&{{a^2} - {b^2}}\\0&{b - c}&{{b^2} - {c^2}}\\1&c&{{c^2}}\end{array}\,} \right|,$     by $\begin{array}{l}{R_1} \to {R_1} - {R_2}\\{R_2} \to {R_2} - {R_3}\end{array}$

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

= $(a - b)\,\,(b - c)\,\left| {\,\begin{array}{*{20}{c}}0&0&{a - c}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}\,} \right|$,       by ${R_1} \to {R_1} - {R_2}$

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

= $(a - b)\,(b - c)\,(a - c)\,.\,( - 1) = (a - b)\,(b - c)\,(c - a)$.

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