Convert Luminous and Color Opponent Channels to RGB Image

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Update - I changed the RGB to opponent conversion to that posted here: https://www.mathworks.com/matlabcentral/answers/118889-rgb-to-opponent-color-space-code

I converted the RGB channels of an image to luminous and color opponent channels; now I am trying to convert the opponent channels back to RGB (using only luminous and the opponent channels), but I do not know how to compute R, G, and B given luminous, red-green, and yellow-blue. How do we go back to RGB from luminuous, red-green, and yellow-blue?

My Method

I converted the RGB image to the opponent channels using the method here. The conversion goes:

//Red minus green
opponentChannels[0].at<float>(y, x) = (r - g) / std::sqrt(2.0f);

//Yellow minus blue
opponentChannels[1].at<float>(y, x) = (r + g - (2 * b)) / std::sqrt(6.0f);

//Luminous
opponentChannels[2].at<float>(y, x) = (r + g + b) / std::sqrt(3.0);

where r, g, and b are floats and x and y are the coodinates of the pixel being set.

To get the original image back, i converted each channel to RGB. For luminous:

cv::Vec3f& pixel = rgbEquivalent->at<cv::Vec3f>(y, x);
//Multiply by sqrt(3.0f) / 3.0f to put in the range of [0.0,1.0].
pixel[0] = pixel[1] = pixel[2] = luminous.at<float>(y, x) * (std::sqrt(3.0f) / 3.0f);

For red-green and yellow-blue (I expect this is where the problem is):

if (type == RED_MINUS_GREEN)
{
    //Multiply by sqrt(2.0f) to put in the range [0.0, 1.0]
    float red = color * std::sqrt(2.0f);
    float green = -red;

    if (red < 0.0f)
        red = 0.0f;

    if (green < 0.0f)
        green = 0.0f;

    pixel[0] = 0.0;
    pixel[1] = green;
    pixel[2] = red;
}
else if (type == YELLOW_MINUS_BLUE)
{
    //Multiply by sqrt(6.0f) / 2.0f to put in the range of [0.0, 1.0]
    float yellow = color * (std::sqrt(6.0f) / 2.0f);
    float blue = -yellow;

    if (yellow < 0.0f)
        yellow = 0.0f;

    if (blue < 0.0f)
        blue = 0.0f;

    pixel[0] = blue;
    pixel[1] = yellow;
    pixel[2] = yellow;
}

For the image:

I achieved these RGB representations of each channel

I then get the original back by taking the average RGB value, but it seems too dark:

Summing all the RGB values results in an image that seems slightly too bright (or yellow?):