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Tensor analysis

  • #1
ε

Homework Statement


show that the determinant of a matrix A can be calculated as followings:

det[A]= 1/6 (A_ii A_jj A_kk + 2 A_ij A_jk A_ki - 3 A_ij A_ji A_kk


Homework Equations





The Attempt at a Solution



use ε_pqr det[A]= ε_ijk A_ip A_jq A_kr

ε_pqr ε_pqr det[A]= ε_ijk ε_pqr A_ip A_jq A_kr

use ε_pqr ε_pqr = 6

det[A]= 1/6 ( ε_ijk ε_pqr A_ip A_jq A_kr)

and ε_ijk ε_pqr = det | δ_ip δ_iq δ_ir |
| δ_jp δ_jq δ_jr |
| δ_kp δ_kq δ_kr |

from here i dont know what to do
and δ - delta and ε - epsilon
 

Answers and Replies

  • #2
vela
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So far so good. Just expand the determinant out now.
$$\begin{vmatrix}
\delta_{ip} & \delta_{iq} & \delta_{ir} \\
\delta_{jp} & \delta_{jq} & \delta_{jr} \\
\delta_{kp} & \delta_{kq} & \delta_{kr}
\end{vmatrix} = \delta_{ip}\delta_{jq}\delta_{kp} + \cdots$$ Then plug the resulting expression into
$$\det(A) = \frac{1}{6}(\varepsilon_{ijk}\varepsilon_{pqr} A_{ip}A_{jq}A_{kr})$$ The Kronecker deltas will allow you to do some of the summations.
 

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