If A is a square matrix of order 3, B = kA and |B| = x|A| then,

x = 2k x = k2 x = k3 x = 3k

x = k3

A → order (3 × 3) B = kA So  |B| = |kA| = k3|A|   (as A has order 3) so given |B| = x|A| x = k3   (upon comparing)