Let $A = [a_{ij}]$ be a square matrix of order 2 with elements either 0 or 1. Then the difference between the possible number of singular and non-singular matrices is

A 2×2 matrix with elements either 0 or 1:

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$

A non-singular matrix has determinant ≠ 0 (invertible).

A singular matrix has determinant = 0 (not invertible).

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Each of the 4 positions can have 2 values (0 or 1).

Total matrices $= 2^4 = 16$

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For a 2×2 matrix, the determinant is:

$\det(A) = a_{11} \cdot a_{22} - a_{12} \cdot a_{21}$

Since each element is either 0 or 1, each product ($a_{11}a_{22}$ and $a_{12}a_{21}$) can only be 0 or 1.

Therefore, $\det(A)$ can only be: $-1$, $0$, or $1$.

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For $\det(A) = 1$:

$a_{11}a_{22} - a_{12}a_{21} = 1$

This requires $a_{11}a_{22} = 1$ and $a_{12}a_{21} = 0$.

Both $a_{11} = 1$ and $a_{22} = 1$, while at least one of $a_{12}$ or $a_{21}$ must be 0.

Possible combinations for $(a_{12}, a_{21})$: $(0, 0)$, $(0, 1)$, $(1, 0)$

Count $= 3$ matrices

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For $\det(A) = -1$:

$a_{11}a_{22} - a_{12}a_{21} = -1$

This requires $a_{12}a_{21} = 1$ and $a_{11}a_{22} = 0$.

Both $a_{12} = 1$ and $a_{21} = 1$, while at least one of $a_{11}$ or $a_{22}$ must be 0.

Possible combinations for $(a_{11}, a_{22})$: $(0, 0)$, $(0, 1)$, $(1, 0)$

Count $= 3$ matrices

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Total non-singular matrices $= 3 + 3 = 6$

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Singular matrices $= 16 - 6 = 10$

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Difference $= 10 - 6 = 4$

Therefore, the answer is $4$.