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3 and 4 .Determinants and Matrices
easy
$\left| {\,\begin{array}{*{20}{c}}1&a&{b + c}\\1&b&{c + a}\\1&c&{a + b}\end{array}\,} \right|= . . .. $
A
$a + b + c$
B
${(a + b + c)^2}$
C
$0$
D
$1 + a + b + c$
Solution
(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&a&{b + c}\\1&b&{c + a}\\1&c&{a + b}\end{array}\,} \right| = (a + b + c)\,\left| {\,\begin{array}{*{20}{c}}1&1&{b + c}\\1&1&{c + a}\\1&1&{a + b}\end{array}\,} \right|$
$({C_2} \to {C_2} + {C_3})= 0$,
$(\because \,\,{{C}_{1}}\equiv {{C}_{2}})$
Standard 12
Mathematics
