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If a matrix has m rows and n columns, then the first row can have 2ⁿ combinations, half of which have an even number of 1's. Then you can fill up the matrix row by row like that, and you do not need to care about the column evenness yet. Then when you come to the last row, you must compensate for what you have done so far, so the last row is given by the previous rows, and since all previous rows have an even number of 1's, the last row will automatically also get an even number of 1's.

Thus, the number of such matrices is:

2 ^{[ (n - 1) × (m - 1) ]}