The absolute value of a $2 \times 2$ matrix determinant is the area of a corresponding parallelogram with the $2$ row vectors as sides.

The absolute value of a $3 \times 3$ matrix determinant is the volume of a corresponding parallelepiped with the $3$ row vectors as sides.

Can it be generalized to $n-D$? The absolute value of an $n \times n$ matrix determinant is the volume of a corresponding $n-$parallelotope?

Martin Sleziak
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Chao Xu
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3 Answers3


Yes it can. In fact, as Jamie Banks noted, a determinant is an intuitive way of thinking about volumes. To summarise the argument, if we consider the vectors as a matrix, switching two rows, multiplying one by a constant or adding a linear combination will have the same effect on the volume as on the determinate. We can use these operations to transform any n-parallelotope to cube and note that the determinate matches the signed volume here, so it will match it everywhere as well.

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Yes--see the Wikipedia article.

This follows from the change of variables formula with the Jacobian. However, it is probably more accurate to say that the change-of-variables formula is a (nontrivial) consequence of this. One argument for this fact, which is given in Rudin's Real and Complex Analysis, is the fact that both the determinant and the volume function up to sign behave very nicely satisfying certain specific properties (i.e., multilinear, alternating, normalized) hence are the same.

Akhil Mathew
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Yes, that is because in its most general form the determinate is defined that way, except without the absolute value part. The determinant is a way to compare volumes of subspaces that have the same grade, which is the multi-vector equivalent of the size of matrix.

Jonathan Fischoff
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