This is dealing with computer vision, but the only part I'm having trouble understanding is a step in the matrix math. So it seems appropriate it should go here.(adsbygoogle = window.adsbygoogle || []).push({});

The paper/chapter I'm reading takes one of those steps saying "from this we can easily derive this" and I'm not quite sure what happened. Perhaps it's just been too long since I've done significant linear algebra.

Basically we have the follow:

The relation from a 3D point to an image point is given by:

[tex]

\left( \begin{array}{ccc} x_1 \\ x_2 \\ x_3 \end{array} \right) = M_{int} M_{ext} \left( \begin{array}{ccc} X_w \\ Y_w \\ Z_w \\ 1 \end{array} \right)

[/tex]

Where:

[tex]

M_{ext} = \left[ \begin{array}{ccc} r_{11} & r_{12} & r_{13} & T_{x} \\ r_{21} & r_{22} & r_{23} & T_{y} \\ r_{31} & r_{32} & r_{33} & T_{z} \end{array} \right]

[/tex]

However, the equation for a plane give us:

[tex]

Z_{w} = d - n_{x} X_{w} - n_{y} Y_{w}

[/tex]

The above equation can be reduced to:

[tex]

\left( \begin{array}{ccc} x_1 \\ x_2 \\ x_3 \end{array} \right) = M_{int} \left[ \begin{array}{ccc} r_{11}-n_{x}r_{13} & r_{12}-n_{y}r_{13} & dr_{13}+T_{x} \\ r_{21}-n_{x}r_{23} & r_{22}-n_{y}r_{23} & dr_{23}+T_{y} \\ r_{31}-n_{x}r_{33} & r_{32}-n_{y}r_{33} & dr_{33}+T_{z} \end{array} \right] \left( \begin{array}{ccc} X_w \\ Y_w \\ 1 \end{array} \right)

[/tex]

So my question is, how is the equation of the plane applied to the 4x3 matrix to reduce to a 3x3 matrix. I can see what happened but I don't understand fully how it happened. If someone could explain that part it would be great.

Thank you for your time!

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# Having trouble understanding a dimension reduction step in a matrix.

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