Converting product of exponentials from base frame to end-effector frame

Question

This is a homework question from edx course Robot Mechanics and Control, Part II

Given the following and expressing its forward kinematics as $T = e^{[S_1]\theta_1} ... e^{[S_6]\theta_6}M$

It is can be found (and also shown in the answer) that

$$ [S_2] = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$ and $$ M = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 3L \\ 0 & 0 & -1 & -2L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

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Part 4b) requires expressing the forward kinematics in the form of $T = Me^{[B_1]\theta_1} ... e^{[B_n]\theta_n}$ and finding $[B_2]$

I wanted to try deriving the answer using the following property (as found in the lecture notes page 20 and in the lecture around 3:00, basically using property $Pe^A = e^{PAP^{-1}}P$ for any invertible $P$):

$$e^{[S_1]\theta_1} ... e^{[S_n]\theta_n}M = Me^{[B_1]\theta_1} ... e^{[B_n]\theta_n}$$ where $$[B_i] = M^{-1}[S_i]M$$

I get $$ M^{-1} = \begin{bmatrix} 0 & 1 & 0 & -3L \\ 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

Using the property $$[B_i] = M^{-1} [S_i] M$$ to calculate $B_2$ and I get $$ [B_2] = \begin{bmatrix} 0 & 0 & 1 & -3L \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -5L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

which is obviously wrong. What am I doing incorrectly?

Thanks in advance

The correct answer is

$$ [B_2] = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -3L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$

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Full question with answers below:

Answer 1

So I finally found my mistake...

The mistake was in the matrix form of the screw

$ [S_2] = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $ should be $ [S_2] = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2L \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $

and I misinterpreted the answer...

$ [B_2] = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -3L \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $ should be $ [B_2] = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -3L \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $

What a fool...