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soumon Let the given determinant be ( Delta ). Then, applying ( C_{3} rightarrow C_{3}-C_{2} ) we get:
[
begin{aligned}
Delta=&left|begin{array}{lll}
a+b & 2 a+b & a
2 a+b & 3 a+b & a
4 a+b & 5 a+b & a
end{array}right|
=left|begin{array}{lll}
a+b & a & a
2 a+b & a & a
4 a+b & a & a
end{array}right| & text { [applying C }_{2} rightarrow C_{2}-C_{1} 1
end{aligned}
]
[
=0
]
( left[because C_{2} text { and } C_{3} ) are identical] right.
Hence, ( Delta=0 ) 
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EXAMPLE 12 Without expanding the determinant, prove that a+b 2a + b 3a + b 2a + b 3a + b 4a + b = 0. 4a + b 5a + b 6a + b
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