Cramer's Rule and Dyadics(Menzel)

  • #1

Main Question or Discussion Point

The following is from Donald H. Menzel's 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.
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
131
Hey Odious Suspect.

This looks like a multi-linear/tensor problem and I think it needs to be pointed out what sort of algebra the i_hat, j_hat and l_hat (looks like a weird l) have.

Are they just the normal cross product relations or are they something else?
 
  • #3
Hey Odious Suspect.

This looks like a multi-linear/tensor problem and I think it needs to be pointed out what sort of algebra the i_hat, j_hat and l_hat (looks like a weird l) have.

Are they just the normal cross product relations or are they something else?
##\hat{\mathfrak{i}}, \hat{\mathfrak{j}}, \hat{\mathfrak{k}}## are the traditional ##\hat{i}, \hat{j}, \hat{k}## of vector calculus. I use German (Fraktur) letters for vector-ish and tensor-ish things. The ##\Phi## beast is a dyadic. I am not comfortable enough with dyadics to provide a "crash course" to get you up to speed. You know what the Uppanishads say about the blind leading the blind.

##\mathfrak{B}_i## are vectors (n-tuples).
 
  • #4
9
3
Hi there. so i usually get problems where you transpose a whole matrix. but we can treat the two rows as a coefficient matrix

if you treat the two rows as a coefficient matrix and transpose them you get [ B1 A11 A12 ]
[ B2 B3 A13 ]

cramer's rule if you want to solve for just one of the variables instead of all is that x1 = det ( A1 ) / det (A) x2 = det (A2) / det (A). You could make B the column 1 variable and solve for B. Transpose the matrix. [ B1 a11 a31 ] det (B) / det ( of the original matrix) = B.
[ B2 a12 a32 ]
[ B3 a13 a33 ]

so what a marvel of a comely matrix problem. hope you have a great day.
 
  • #5
Hi there. so i usually get problems where you transpose a whole matrix. but we can treat the two rows as a coefficient matrix

if you treat the two rows as a coefficient matrix and transpose them you get [ B1 A11 A12 ]
[ B2 B3 A13 ]

cramer's rule if you want to solve for just one of the variables instead of all is that x1 = det ( A1 ) / det (A) x2 = det (A2) / det (A). You could make B the column 1 variable and solve for B. Transpose the matrix. [ B1 a11 a31 ] det (B) / det ( of the original matrix) = B.
[ B2 a12 a32 ]
[ B3 a13 a33 ]

so what a marvel of a comely matrix problem. hope you have a great day.
Thank you for taking the time to read and reply. My question has to do with the ordering of rows in the last application of Cramer's rule in the original post.

##\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|} ##

The replacement of rows in the determinant of the numerator is appropriate due to the fact (which I didn't emphasize) that the coefficients in the system of simultaneous linear equations are those of the transposed matrix. The problem is to solve (invert) the following equation:

##\left[\begin{array}{ccc} \hat{\mathfrak{i}} & \hat{\mathfrak{j}} & \hat{\mathfrak{k}}\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]=\left[\begin{array}{ccc} \mathfrak{B}_{1} & \mathfrak{B}_{2} & \mathfrak{B}_{2}\end{array}\right]##

It appears to me that Menzel should have written

##\mathfrak{j}=\frac{\left| \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
\mathfrak{B}_1 & \mathfrak{B}_2 & \mathfrak{B}_3 \\
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|} ##
 

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