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Explanation:

Total number of matrices formed = Number of ways of selecting 4 distinct integer × Arranging these 4 integers in the matrix.
= ⁶C₄ × 4! = 15 × 24 = 360

For a singular matrix $\left|\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right|$, its determinant = 0.
∴ ad - bc = 0
⇒ ad = bc

Now, possible pairs of a and d can be

Case 1: a and b are 2 or 3, 
∴ a × d = b × c = 6
⇒ b and c are 1 or 6.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 2: a and b are 1 or 6,
∴ a × d = b × c = 6
⇒ b and c are 2 or 3.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 3: a and b are 3 or 4, 
∴ a × d = b × c = 12
⇒ b and c are 2 or 6.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 4: a and b are 2 or 6, 
∴ a × d = b × c = 12
⇒ b and c are 3 or 4.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

⇒ Total number of required matrices = 4 + 4 + 4 + 4 = 16

Required probability = 16/360 = 2/45.

Hence, option (a).