3 and 4 .Determinants and Matrices
medium

જો $a,b,c$ એ અસમાન હોય તો $\Delta = \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} + 1}\\b&{{b^2}}&{{b^3} + 1}\\c&{{c^2}}&{{c^3} + 1}\end{array}\,} \right|= 0$ માટે . . . .શરતનું પાલન થવું જોઈએ.

A

$1 + abc = 0$

B

$a + b + c + 1 = 0$

C

$(a - b)(b - c)(c - a) = 0$

D

એકપણ નહી.

(IIT-1985)

Solution

(a) Splitting the determinant into two determinants, we get

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

= $(1 + abc)\,[(a – b)\,(b – c)\,(c – a)] = 0$

Because $a, b, c$ are different, the second factor cannot be zero.

Hence, option $(a)$, $1 + abc = 0$, is correct.

Standard 12
Mathematics

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