What does $\ cof(M)[t]_×$ equal to?

Question

Say $\ cof(M)[\mathbf t]_×=[M\mathbf t]_×M$,

What does $\ [\mathbf t]_×cof(M)$ equal to?

And What about $\ [\mathbf t]_×M$ ?

M is 3 x 3 matrix, $\ cof()$means cofactor matrix,t is 3x1 vector,$\ [ \ ]_×$means convert cross product to matrix form.

Answer 1

$$\ [\mathbf t]_×cof(M)=\begin{vmatrix} 0 & -t_z & t_y \\ t_z & 0 & -t_x \\ -t_y & t_x & 0 \\ \end{vmatrix}*\begin{vmatrix} M^c_{11}&M^c_{12}&M^c_{13}\\M^c_{21}&M^c_{22}&M^c_{23}\\M^c_{31}&M^c_{32}&M^c_{33}\end{vmatrix}=\begin{Bmatrix} -M^c_{21}t_z+M^c_{31}t_y & -M^c_{22}t_z+M^c_{32}t_y & -M^c_{23}t_z+M^c_{33}t_y\\ M^c_{11}t_z-M^c_{31}t_x & M^c_{12}t_z-M^c_{32}t_x & M^c_{13}t_z-M^c_{33}t_x\\ -M^c_{11}t_y+M^c_{21}t_x & -M^c_{12}t_y+M^c_{22}t_x & -M^c_{13}t_y+M^c_{23}t_x \end{Bmatrix}=\left\{\begin{array}{l} (M_{12}M_{33}-M_{13}M_{32})t_z+(M_{12}M_{23}-M_{13}M_{22})t_y \\ (M_{22}M_{33}-M_{23}M_{32})t_z+(M_{13}M_{22}-M_{12}M_{23})t_x \\ (M_{23}M_{32}-M_{22}M_{33})t_y+(M_{13}M_{32}-M_{12}M_{33})t_x \end{array}\right. \left.\begin{array}{l} (M_{13}M_{31}-M_{11}M_{33})t_z+(M_{13}M_{21}-M_{11}M_{23})t_y\\ (M_{23}M_{31}-M_{21}M_{33})t_z+(M_{11}M_{23}-M_{13}M_{21})t_x\\ (M_{21}M_{33}-M_{23}M_{31})t_y+(M_{11}M_{33}-M_{13}M_{31})t_x \end{array}\right. \left.\\\begin{array}{l} (M_{11}M_{32}-M_{12}M_{31})t_z+(M_{11}M_{22}-M_{12}M_{21})t_y\\ (M_{21}M_{32}-M_{22}M_{31})t_z+(M_{12}M_{21}-M_{11}M_{22})t_x\\ (M_{22}M_{31}-M_{21}M_{32})t_y+(M_{12}M_{31}-M_{11}M_{32})t_x \end{array}\right\}\\=\begin{Bmatrix} M_{11}&M_{12}&M_{13}\\M_{21}&M_{22}&M_{23}\\M_{31}&M_{32}&M_{33} \end{Bmatrix}*\\\begin{Bmatrix}\begin{array}{ccc} 0 & -(M_{13}t_x+M_{23}t_y+M_{33}t_z) & M_{12}t_x+M_{22}t_y+M_{32}t_z\\ M_{13}t_x+M_{23}t_y+M_{33}t_z & 0 & -(M_{11}t_x+M_{21}t_y+M_{31}t_z)\\ -(M_{12}t_x+M_{22}t_y+M_{32})t_z & M_{11}t_x+M_{21}t_y+M_{31}t_z & 0 \end{array}\end{Bmatrix}=M[M^Tt]_×$$ Therefore

$\ [\mathbf t]_×cof(M)=M[M^Tt]_×$

Answer 2

$\ M^*$ is adjugate matrix of $\ M, M^c$ is cofactor matrix of $\ M$ $$\ M^-=\frac {M^*}{|M|} \Rightarrow M^{-T}=\frac {M^c}{|M|}\Rightarrow M^{T}=\frac{|M|}{M^c}\Rightarrow \frac {M^T}{|M|}=\frac{1}{M^c}$$ Becasue M is of rank 3,therefore $$\\ \Biggl|\frac {M^t}{|M|}\Biggr|=\Biggl|\frac {|M^t|}{|M|^3}\Biggr|=\frac {1}{|M|^2}=\frac {1}{|M^c|}$$ therefore $$\ |M|=\pm|M^c|^{\frac{1}{2}}$$ then$$\ M^T=\pm\frac{|M^c|^{\frac{1}{2}}}{M^c} \ \ M=\pm\frac{|M^{c}|^{\frac{1}{2}}}{M^{cT}}$$ replace $\ M\ M^T\ with\ M^c$ $$\\ [\mathbf t]_×M^c=M[M^T\mathbf t]_×=\frac{|M^c|^{\frac{1}{2}}}{M^{cT}}\biggl[\frac{|M^{c}|^{\frac{1}{2}}}{M^{c}}\mathbf t\biggr]_×=|M^c|\cdot M^{-cT}\cdot[M^{c-}\mathbf t]_×$$ Because gernerality of this equation we can rewrite $\ M^c$ as $\ M $

$$\\ [\mathbf t]_×M=|M|\cdot M^{-T}\cdot[M^{-}\mathbf t]_×$$