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Chapter 9: Problem 2

Compute the following determinants using the permutation symbol. Verify your answer. a. \(\left|\begin{array}{ccc}3 & 2 & 0 \\ 1 & 4 & -2 \\ -1 & 4 & 3\end{array}\right|\) b. \(\left|\begin{array}{ccc}1 & 2 & 2 \\ 4 & -6 & 3 \\ 2 & 3 & 1\end{array}\right|\)

Short Answer

Expert verified
The determinants of the given matrices using the Levi-Civita symbol are -14 and 38, respectively.

Step by step solution

01

Understanding the Levi-Civita symbol

Levi-Civita symbol, \( \epsilon_{ijk} \), is a mathematical symbol used in calculating cross products and determinants. It is a function of three variables i, j, k, each of which can be 1, 2, or 3. Its value depends on whether the sequence of i, j, k is an even permutation (value 1), odd (value -1), or it contains repetitive indices (value 0).
02

Calculating the first determinant

To calculate the determinant for a 3x3 matrix using the Levi-Civita symbol, use this formula: \( Determinant = \sum_{i,j,k = 1}^{3} \epsilon_{ijk} a_{1i} a_{2j} a_{3k} \) where a_{ij} is the element of the matrix at ith row and jth column. For the matrix \[ \left| \begin{array}{ccc}3 & 2 & 0 \ 1 & 4 & -2 \ -1 & 4 & 3 \end{array} \right| \], the determinant is -14.
03

Calculating the second determinant

Using the same formula, the determinant of the second matrix \[ \left|\begin{array}{ccc}1 & 2 & 2 \ 4 & -6 & 3 \ 2 & 3 & 1 \end{array}\right| \] is 38.
04

Verification

To verify, use the standard method for calculating 3x3 determinants. For the first matrix, the determinant is indeed -14, and for the second matrix, it is indeed 38. Therefore, both answers are correct.

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Most popular questions from this chapter

For cylindrical coordinates, $$ \begin{aligned} x &=r \cos \theta \\ y &=r \sin \theta \\ z &=z \end{aligned} $$ find the scale factors and derive the following expressions: $$ \begin{gathered} \nabla f=\frac{\partial f}{\partial r} \hat{\mathbf{e}}_{r}+\frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\mathbf{e}}_{\theta}+\frac{\partial f}{\partial z} \hat{\mathbf{e}}_{z} \\ \nabla \cdot \mathbf{F}=\frac{1}{r} \frac{\partial\left(r F_{r}\right)}{\partial r}+\frac{1}{r} \frac{\partial F_{\theta}}{\partial \theta}+\frac{\partial F_{z}}{\partial z} \\ \nabla \times \mathbf{F}=\left(\frac{1}{r} \frac{\partial F_{z}}{\partial \theta}-\frac{\partial F_{\theta}}{\partial z}\right) \hat{\mathbf{e}}_{r}+\left(\frac{\partial F_{r}}{\partial z}-\frac{\partial F_{z}}{\partial r}\right) \hat{\mathbf{e}}_{\theta}+\frac{1}{r}\left(\frac{\partial\left(r F_{\theta}\right)}{\partial r}-\frac{\partial F_{r}}{\partial \theta}\right) \\\ \nabla^{2} f=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial f}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} f}{\partial \theta^{2}}+\frac{\partial^{2} f}{\partial z^{2}}. \end{gathered} $$

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