Consider $4 \times 4$ matrices with their elements from $\{0, 1\}$. The number of such matrices with even number of $1$s in every row and every column is

Let the matrix $A$ be of size $n \times n$ with $n = 4$ and entries in $\{0, 1\}$.

The condition that each row and each column contains an even number of $1$s is equivalent to saying that the sum of entries in each row and each column is $0 \pmod 2$.

1. For the upper-left $(n-1) \times (n-1)$ submatrix (i.e., size $3 \times 3$), each entry can be chosen independently as either $0$ or $1$. There are $(n-1)^2 = 3^2 = 9$ entries, giving $2^9 = 512$ ways.

2. Once the first $n-1$ entries of the first $n-1$ rows are chosen:

   • The last element of each of the first $n-1$ rows (i.e., $A_{i, 4}$ for $i=1,2,3$) is uniquely determined by the parity requirement of row $i$.

   • The last element of each of the first $n-1$ columns (i.e., $A_{4, j}$ for $j=1,2,3$) is uniquely determined by the parity requirement of column $j$.

3. The bottom-right entry $A_{4,4}$ is then uniquely determined to satisfy the parity of the 4th row (which automatically satisfies the parity of the 4th column as well, since the sum of all elements modulo 2 must be consistent).

Therefore, the total number of valid matrices is:

Hence, the correct option is (A).