The determinant of a matrix

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Tensors and differential geometry · Tensors and differential geometry

Invariance, Ch Differential forms and the exterior derivative

Problem

The determinant of a 3×33\times3 matrix AijA_{ij}, as an index formula in Cartesian tensors, is det⁡A=ϵijkA1iA2jA3k\det A = \epsilon_{ijk}A_{1i}A_{2j}A_{3k}. Consider the three one-forms α1=2 dx+dy−dz,α2=dx+3 dy+dz,α3=dx−dy+2 dz.\begin{aligned}\alpha^1&=2\,\mathrm{d}x+\mathrm{d}y-\mathrm{d}z,\\ \alpha^2&=\mathrm{d}x+3\,\mathrm{d}y+\mathrm{d}z,\\ \alpha^3&=\mathrm{d}x-\mathrm{d}y+2\,\mathrm{d}z.\end{aligned} ((a)) Use the identity α1∧α2∧α3=det⁡(αai) dx∧dy∧dz\alpha^1\wedge\alpha^2\wedge\alpha^3=\det(\alpha^a{}_i)\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z, where αai\alpha^a{}_i is the matrix of coefficients (row aa, column i=x,y,zi=x,y,z), to compute α1∧α2∧α3\alpha^1\wedge\alpha^2\wedge\alpha^3. ((b)) Confirm the coefficient by evaluating det⁡A\det A with the index formula ϵijkA1iA2jA3k\epsilon_{ijk}A_{1i}A_{2j}A_{3k}, and check the two results agree.

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