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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&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right| = $
  • A
    ${a^3} + {b^3} + {c^3} - 3abc$
  • B
    ${a^3} + {b^3} + {c^3} + 3abc$
  • ✓
    $(a + b + c)(a - b)(b - c)(c - a)$
  • D
    None of these
✓

Answer

Correct option: C.
$(a + b + c)(a - b)(b - c)(c - a)$
c
(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|$ vanishes when $a = b,\,b = c,\,c = a$.

Hence $(a - b),\,(b - c),\,(c - a)$ are factors of $\Delta $. Since $\Delta $ is symmetric in $a,b,c $ and of $4th$  degree, $(a + b + c)$ is also a factor, so that we can write

$\Delta$=$k(a-b)(b-c)(c-a)(a+b+c)  $     ......................$(i)$

Where by comparing the coefficients of the leading term $b{c^3}$ on both the sides of identity   $(i).$ We get $1 = k( - 1)\,( - 1) \Rightarrow k = 1$

$\Delta$=$(a-b)(b-c)(c-a)(a+b+c)    $

Trick : Put $a = 1,\,b = 2,\,c = 3$, so that determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&8&{27}\end{array}\,} \right| = 1(30) - 1(24) + 1(8 - 2) = 12$

which is given by $(c)$ . i.e. $(1 + 2 + 3)\,(1 - 2)\,(2 - 3)(3 - 1) = 12$.

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