Applying R₁→R₁ + R₂ + R₃, we get,

Taking, (3a + 2b) common we get,

Applying, C₁→C₁ – C₂ and C₃→C₃ – C₂, we get,

= 3(a + b)b²(3) = 9(a + b) b²

= R.H.S

Hence, proved.