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Using properties of determinants, prove the following: |(a, a + b, a + 2b), (a + 2b, a, a + b), (a + b, a + 2b, a)| = 9b2(a + b)

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Applying R1  R1 + R2 + R3, we have

Taking out 3(a + b) from 1st row, we have

Applying C1 → C1 – C2 and C2 → C2 – C3 

Expanding along first row, we have 

D = 3(a + b) [1. (4b2 – b2)] 

= 3(a + b) x 3b2 = 9b2(a + b)

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