The following is from Donald H. Menzel's Mathematical Physics:(adsbygoogle = window.adsbygoogle || []).push({});

##\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.

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# I Cramer's Rule and Dyadics(Menzel)

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