# Finding translation component of a 4x4 matrix inverse

I have 4x4 matrix (A) (maya softw):

|Xx Xy Xz 0| - basis of X and dummy 0
|Yx Yy Yz 0| - basis of Y and dummy 0
|Zx Zy Zz 0| - basis of Z and dummy 0
|Tx Ty Tz 1| - Position x, y, z and dummy 1


In this case, I know there is no scaling or sheer, so the upper-left 3x3 portion is a pure rotation matrix. That means I can invert this portion by taking the transpose.

So inverting the matrix should give me something like this:

|Xx Yx Zx 0|
|Xy Yy Zy 0|
|Xz Yz Zz 0|
|T? T? T? 1|


But I don't know how to find the T? values that correctly invert the translation component of the matrix. It's not just the inverted position (T-1)

My thinking so far:

Matrix of an untransformed object at the origin - (I - identity):

| 1  0  0  0 |
| 0  1  0  0 |
| 0  0  1  0 |
| 0  0  0  1 |


Matrix of a translated object (Tx=3, Ty=4, Tz=5) (A):

| 1  0  0  0 |
| 0  1  0  0 |
| 0  0  1  0 |
| 3  4  5  1 |


Inverse matrix of A (A'):

| 1  0  0  0 |
| 0  1  0  0 |
| 0  0  1  0 |
|-3 -4 -5  1 |


Matrix of translated and rotated(Y=90) object(B):

| 0  0 -1  0 |
| 0  1  0  0 |
| 1  0  0  0 |
| 3  4  5  1 |


inverse matrix of B (B'):

| 0  0  1  0 |
| 0  1  0  0 |
|-1  0  0  0 |
| 5 -4 -3  1 |



Strange things start to happen when you take two differently-rotated matrices with the same translation and you get different inverse translations!

It seems like some dot product is involved in the translation calculation.

• how can I temporally hide this post.. I found some solution.. but don't exactly confident with that – bafly Mar 9 at 16:12
• If you found a solution, post it as an Answer. I don't recommend deleting the post, as that would prevent users from offering you answers that might give you more confidence. – DMGregory Mar 9 at 16:40
• Once a post is deleted, I think the system allows you to undelete it for a period of time. – Vaillancourt Mar 9 at 16:46

If I have a matrix $$\M\$$ with translation vector $$\\vec T\$$, that means it maps the origin to the vector $$\\vec T\$$ in homogeneous coordinates, ie:

$$\begin{bmatrix}0 & 0 & 0 & 1\end{bmatrix} \cdot M = \begin{bmatrix}T_x & T_y & T_z & 1 \end{bmatrix} = \begin{bmatrix} \vec T & 1 \end{bmatrix}$$

So then the inverse of $$\M\$$, that is $$\M^{-1}\$$, should do the opposite, and map $$\ \vec T\$$ back to the origin.

We can think of the result of the matrix multiplication as a sum of the input vector multiplied by the first three rows (what the rotation / scaling / shearing part of the matrix do to the vector) plus the input vector multiplied by the final row (what the translation part of the matrix does to the vector). To get back to zero, this last part has to exactly negate what happened before.

So, if we look at just the upper 3x3 portion of your given inverse does to the vector $$\\vec T\$$...

$$\begin{bmatrix}T_x & T_y & T_z\end{bmatrix} \cdot \begin{bmatrix}X_x & Y_x & Z_x \\ X_y & Y_y & Z_y\\X_z & Y_z & Z_z \end{bmatrix} = \begin{bmatrix} \vec T \cdot \vec X & \vec T \cdot \vec Y & \vec T \cdot \vec Z\end{bmatrix}$$

This gives us what our inverse translation needs to undo. So our final translation vector, call it $$\\vec U\$$ for "Undo", is:

$$\vec U = -1 \cdot\begin{bmatrix} \vec T \cdot \vec X & \vec T \cdot \vec Y & \vec T \cdot \vec Z\end{bmatrix}$$

And your inverted 4x4 matrix is:

$$\begin{bmatrix}X_x & Y_x & Z_x & 0\\ X_y & Y_y & Z_y & 0 \\X_z & Y_z & Z_z & 0 \\ -\vec T \cdot \vec X & -\vec T \cdot \vec Y & -\vec T \cdot \vec Z & 1\end{bmatrix}$$

• oh.. is this dot products? - [ T ⃗ ⋅ X ⃗ T ⃗ ⋅ Y ⃗ T ⃗ ⋅ Z ⃗ ] – bafly Mar 9 at 20:42
• It's a product written with a dot, so it looks like it fits the bill. ;) Is there anything that would lead you to believe it's not a dot product? – DMGregory Mar 9 at 20:43
• ok.. give me some time.. I must chew all that stuff.. my brains are melted.. or just do straight ahead what you've wrote in my code.. and watch that mathemagic start on my screen.. thank you very very much for your time! – bafly Mar 9 at 20:48
• oh! and thank you for your grammar and logical edits! – bafly Mar 9 at 20:49
• wow, its working! thx DMGregory! so I need to keep matrix clean from sheers and scales to keep this working! – bafly Mar 10 at 0:06