If , then show that .

Expanding along the first row

= 1(1×4 – 2×0) – 0(0×4 – 0×2) + 1(0×0 – 0×1)

= 1(4 – 0) + 0 + 1(0 + 0)

= 1×4

= 4

Now

Expanding along the first row

= 3(3×12 – 6×0) – 0(0×12 – 0×6) + 3(0×0 – 0×3)

= 3(36 – 0) + 0 + 3(0 + 0)

= 3×36

= 108

= 27 × 4

= 27 |A|

Hence, |3A|= 27 |A|