Question

sourtion We have
[
begin{aligned}
L H S &=left|begin{array}{lll}
a+b & b+c & c+a
b+c & c+a & a+b
c+a & a+b & b+c
end{array}right|
&=left|begin{array}{ccc}
-2 c & b+c & c+a
-2 a & c+a & a+b
-2 b & a+b & b+c
end{array}right| text { [applying } C_{1} rightarrow C_{1}-left(C_{2}+C_{3}right)
end{aligned}
]
( left.=(-2)left|begin{array}{ccc}c & b+c & c+a a & c+a & a+b b & a+b & b+cend{array}right| text { [taking out }(-2) text { common from } C_{1}right] )
( left.=(-2) cdotleft|begin{array}{lll}c & b & a a & c & b b & a & cend{array}right| text { [applying } C_{2} rightarrow C_{2}-C_{1} text { and } C_{3} rightarrow C_{3}-C_{1}right] )
[
begin{aligned}
&=2left|begin{array}{lll}
a & b & c
b & c & a
c & a & b
end{array}right| &left[text { applying } C_{1} leftrightarrow C_{3}right]
=& text { RHS. }
end{aligned}
]
Hence, ( mathrm{LHS}=mathrm{RHS} )

# EXAMPLE 39 Using the properties of determinants, prove that la +b btc cra la b c b+c cta a+b=2 b c a. lcra a+b b+c le a b

Solution