If then show that |3A| = 27|A|

We know that a determinant of a 3 x 3 matrix is calculated as

= 1[1(4) - 2(0)] – 0 + 1[0-0]

= 1[4 - 0] – 0 + 0

= 4

|A|= 4

LHS: |3A|

= 3[3(12) - 0(6)] – 0 + 3[0 – 0]

= 3(36) – 0 + 0

= 108

|3A| = 108 ----LHS

RHS: 27|A|

27|A| = 27(4)

= 108

27|A| = 108 ----RHS

LHS=RHS

Hence proved.