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This is a solution without STL. It is very ugly, though.

// keep only unique rows
Mat allPairs;       // Mat with duplicate values
Mat uniqueStrides;  // Mat that will contain the unique values
uniqueStrides.push_back( allPairs.row(0) );
for (int i = 1; i < allPairs.rows; ++i) {
    int isInside = false;
    for (int j = 0; j < uniqueStrides.rows; ++j) {
        int count = 0;
        for (int k = 0; k < uniqueStrides.cols; ++k) // checks by element of 
            if(allPairs.at<int>(i,k) == uniqueStrides.at<int>(j,k)) 
                ++count;
        if (count == 2) {
            isInside = true;
            break;
        }   
    }
    if (isInside == false) uniqueStrides.push_back( allPairs.row(i) );
}

This is a solution without STL. It is very ugly, though.

// keep only unique rows
Mat allPairs;       // Mat with duplicate values
Mat uniqueStrides;  // Mat that will contain the unique values
uniqueStrides.push_back( allPairs.row(0) );
for (int i = 1; i < allPairs.rows; ++i) {
    int isInside = false;
    for (int j = 0; j < uniqueStrides.rows; ++j) {
        int count = 0;
        for (int k = 0; k < uniqueStrides.cols; ++k) // checks by element of 
            if(allPairs.at<int>(i,k) == uniqueStrides.at<int>(j,k)) 
                ++count;
        if (count == 2) {
            isInside = true;
            break;
        }   
    }
    if (isInside == false) uniqueStrides.push_back( allPairs.row(i) );
}

This algorithm checks for every uniqueStrides entry and for every value. As my matrices are n x 2, I check if the number of equal values is two:

...
if (count == 2)
    isInside = true;
...

If positive, then this row is already in my matrix.