Given this transformation matrix, how do I decompose it into translation, rotation and scale matrices? I have this problem from my Graphics course. Given this transformation matrix:
$$\begin{pmatrix}
-2 &-1&  2\\
-2  &1& -1\\
 0  &0&  1\\
\end{pmatrix}$$
I need to extract translation, rotation and scale matrices.
I've also have the answer (which is $TRS$):
$$T=\begin{pmatrix}
1&0&2\\
0&1&-1\\
0&0&1\end{pmatrix}\\
R=\begin{pmatrix}
1/\sqrt2 & -1/\sqrt2 &0 \\
1/\sqrt2 & 1/\sqrt2 &0 \\
0&0&1
\end{pmatrix}\\
S=\begin{pmatrix}
-2/\sqrt2 & 0 & 0 \\
0 & \sqrt2 & 0 \\
0& 0& 1
\end{pmatrix}
%    1 0  2        1/sqrt(2) -1/sqrt(2) 0         -2/sqrt(2) 0     0
%T = 0 1 -1    R = /1/sqrt(2) 1/sqrt(2) 0     S = 0        sqrt(2) 0
%    0 0  1        0          0         1         0          0     1
$$
I just have no idea (except for the Translation matrix) how I would get to this solution.
 A: I am a person from the future, and I had the same problem.  For future reference, here's the algorithm for 4x4.  You can solve your 3x3 problem by padding out your problem to the larger dimensions.
Caveat: the following only works for a matrix containing rotation, translation, and nonnegative scalings.  This is the overwhelmingly commonest case, and doubtless what OP was expected to assume.  A more-general solution is significantly more complicated and was not provided because OP didn't ask for it; anyone interested in supporting e.g. negative scalings and shear should look at Graphics Gems II §VII.1.
Start with a transformation matrix:$$
 \begin{bmatrix}
  a & b & c & d\\ 
  e & f & g & h\\ 
  i & j & k & l\\ 
  0 & 0 & 0 & 1
 \end{bmatrix}
$$

*

*Extract Translation
This is basically the last column of the matrix:$$
 \vec{t} = \langle ~d,~h,~l~ \rangle
$$While you're at it, zero them in the matrix.


*Extract Scale
For this, take the length of the first three column vectors:$$
 s_x = \|\langle ~a,~e,~i~ \rangle\|\\
 s_y = \|\langle ~b,~f,~j~ \rangle\|\\
 s_z = \|\langle ~c,~g,~k~ \rangle\|\\
 \vec{s} = \langle s_x,s_y,s_z \rangle
$$


*Extract Rotation
Divide the first three column vectors by the scaling factors you just found.  Your matrix should now look like this (remember we zeroed the translation):$$
 \begin{bmatrix}
  a/s_x & b/s_y & c/s_z & 0\\ 
  e/s_x & f/s_y & g/s_z & 0\\ 
  i/s_x & j/s_y & k/s_z & 0\\ 
  0 & 0 & 0 & 1
 \end{bmatrix}
$$This is the rotation matrix.  There are methods to convert it to quaternions, and from there to axis-angle, if you want either of those instead.
resource
A: It appears you are working with Affine Transformation Matrices, which is also the case in the other answer you referenced, which is standard for working with 2D computer graphics.  The only difference between the matrices here and those in the other answer is that yours use the square form, rather than a rectangular augmented form.
So, using the labels from the other answer, you would have
$$
\left[
\begin{array}{ccc}
a & b & t_x\\
c & d & t_y\\
0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}
s_{x}\cos\psi & -s_{x}\sin\psi & t_x\\
s_{y}\sin\psi & s_{y}\cos\psi & t_y\\
0 & 0 & 1\end{array}\right]
$$
The matrices you seek then take the form:
$$
T=\begin{pmatrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1 \end{pmatrix}\\
R=\begin{pmatrix}
\cos{\psi} & -\sin{\psi} &0 \\
\sin{\psi} & \cos{\psi} &0 \\
0 & 0 & 1 \end{pmatrix}\\
S=\begin{pmatrix}
s_x & 0 & 0 \\
0 & s_y & 0 \\
0 & 0 & 1 \end{pmatrix}
$$
If you need help with extracting those values, the other answer has explicit formulae.
