left| {,begin{array}{*{20}{c}}1&1&1a& | vedclass

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Question

(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&1&1\a&b&c\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} ight|$ vanishes when $a =

Vedclass · Accepted Answer

$(a + b + c)(a - b)(b - c)(c - a)$

Vedclass · Answer

(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&1&1\a&b&c\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} ight|$ 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 $\left| {\,\begin{array}{*{20}{c}}1&1&1\1&2&3\1&8&{27}\end{array}\,} ight| = 1(30) - 1(24) + 1(8 - 2) = 12$

which is given by $(c)$ . i.e. $(1 + 2 + 3)\,(1 - 2)\,(2 - 3)(3 - 1) = 12$.

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

Similar Questions

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