- Home
- Standard 12
- Mathematics
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