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## Main Question or Discussion Point

The following is from Donald H. Menzel's

##\Phi =\left(

\begin{array}{c}

a_{11}\hat{\mathfrak{i}} \hat{\mathfrak{j}}

+a_{12}\hat{\mathfrak{i}} \hat{\mathfrak{j}}

+a_{13}\hat{\mathfrak{i}} \hat{\mathfrak{k}} \\

+a_{21}\hat{\mathfrak{j}} \hat{\mathfrak{i}}

+a_{22}\hat{\mathfrak{j}} \hat{\mathfrak{j}}

+a_{23}\hat{\mathfrak{j}} \hat{\mathfrak{k}} \\

+a_{31}\hat{\mathfrak{k}} \hat{\mathfrak{i}}

+a_{32}\hat{\mathfrak{k}} \hat{\mathfrak{j}}

+a_{33}\hat{\mathfrak{k}} \hat{\mathfrak{k}} \\

\end{array}

\right)=\hat{\mathfrak{i}} \mathfrak{B}_1+\hat{\mathfrak{j}} \mathfrak{B}_2+\hat{\mathfrak{k}} \mathfrak{B}_3##

##\mathfrak{B}_1=\hat{\mathfrak{i}} a_{11}+\hat{\mathfrak{j}} a_{21}+\hat{\mathfrak{k}} a_{31}=\Phi \cdot \hat{\mathfrak{i}}##

##\mathfrak{B}_2=\hat{\mathfrak{i}} a_{12}+\hat{\mathfrak{j}} a_{22}+\hat{\mathfrak{k}} a_{32}=\Phi \cdot \hat{\mathfrak{j}}##

##\mathfrak{B}_3=\hat{\mathfrak{i}} a_{13}+\hat{\mathfrak{j}} a_{23}+\hat{\mathfrak{k}} a_{33}=\Phi \cdot \hat{\mathfrak{k}}##

##\mathfrak{i}=\frac{\left| \begin{array}{ccc}

\mathfrak{B}_1 & \mathfrak{B}_2 & \mathfrak{B}_3 \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33} \\

\end{array}\right|

}{

\left|

\begin{array}{ccc}

a_{11} & a_{12} & a_{13} \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33} \\

\end{array}

\right|} ##

##\mathfrak{j}=\frac{\left| \begin{array}{ccc}

\mathfrak{B}_1 & \mathfrak{B}_2 & \mathfrak{B}_3 \\

a_{11} & a_{12} & a_{13} \\

a_{31} & a_{32} & a_{33} \\

\end{array} \right|

}{

\left|

\begin{array}{ccc}

a_{11} & a_{12} & a_{13} \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33} \\

\end{array}

\right|} ##

Should the first and second rows be transposed in the numerator of the last equation? It appears that the expression, as given, will result in the negative of the advertised value.

__Mathematical Physics__:##\Phi =\left(

\begin{array}{c}

a_{11}\hat{\mathfrak{i}} \hat{\mathfrak{j}}

+a_{12}\hat{\mathfrak{i}} \hat{\mathfrak{j}}

+a_{13}\hat{\mathfrak{i}} \hat{\mathfrak{k}} \\

+a_{21}\hat{\mathfrak{j}} \hat{\mathfrak{i}}

+a_{22}\hat{\mathfrak{j}} \hat{\mathfrak{j}}

+a_{23}\hat{\mathfrak{j}} \hat{\mathfrak{k}} \\

+a_{31}\hat{\mathfrak{k}} \hat{\mathfrak{i}}

+a_{32}\hat{\mathfrak{k}} \hat{\mathfrak{j}}

+a_{33}\hat{\mathfrak{k}} \hat{\mathfrak{k}} \\

\end{array}

\right)=\hat{\mathfrak{i}} \mathfrak{B}_1+\hat{\mathfrak{j}} \mathfrak{B}_2+\hat{\mathfrak{k}} \mathfrak{B}_3##

##\mathfrak{B}_1=\hat{\mathfrak{i}} a_{11}+\hat{\mathfrak{j}} a_{21}+\hat{\mathfrak{k}} a_{31}=\Phi \cdot \hat{\mathfrak{i}}##

##\mathfrak{B}_2=\hat{\mathfrak{i}} a_{12}+\hat{\mathfrak{j}} a_{22}+\hat{\mathfrak{k}} a_{32}=\Phi \cdot \hat{\mathfrak{j}}##

##\mathfrak{B}_3=\hat{\mathfrak{i}} a_{13}+\hat{\mathfrak{j}} a_{23}+\hat{\mathfrak{k}} a_{33}=\Phi \cdot \hat{\mathfrak{k}}##

##\mathfrak{i}=\frac{\left| \begin{array}{ccc}

\mathfrak{B}_1 & \mathfrak{B}_2 & \mathfrak{B}_3 \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33} \\

\end{array}\right|

}{

\left|

\begin{array}{ccc}

a_{11} & a_{12} & a_{13} \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33} \\

\end{array}

\right|} ##

##\mathfrak{j}=\frac{\left| \begin{array}{ccc}

\mathfrak{B}_1 & \mathfrak{B}_2 & \mathfrak{B}_3 \\

a_{11} & a_{12} & a_{13} \\

a_{31} & a_{32} & a_{33} \\

\end{array} \right|

}{

\left|

\begin{array}{ccc}

a_{11} & a_{12} & a_{13} \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33} \\

\end{array}

\right|} ##

Should the first and second rows be transposed in the numerator of the last equation? It appears that the expression, as given, will result in the negative of the advertised value.