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I'm now trying to understand determinants of linear transformations, and I was introduced to determinants as (the scaling factor that we multiply it by any area to figure out what the area of the image of the first area is going to be under a transformation), and I was told that the same thing goes when dealing with 3-dimensional transformations (but this time with volumes). So my question is can we extend that relation between determinants and shapes to further (4,5,6,...,etc.) dimensions? and, if we can, then how can we relate the formula of $n \times n$ matrices’ determinants $$\det(A)=\sum_{j=1}^n (-1)^{i+j} a_{ij}\det(A_{ij})$$(where $j$ is the column (unfixed) and $i$ is the row (fixed) and $A_{ij}$ is the $n-1 \times n-1$ matrix that remains after removing the $i^{th}$ row and the $j^{th}$ column)?

Pietro Paparella
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Khaled Oqab
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    Yes, the analogy works in any dimension. But what is “the equation of determinants”? – David K Jan 27 '19 at 14:49
  • oh sorry i've edited the question and added the formula – Khaled Oqab Jan 27 '19 at 16:12
  • OK, that is one of the formulas to compute a determinant. Are you looking for a geometric interpretation of this particular formula? – David K Jan 27 '19 at 22:48
  • Yes that's exactly what I'm looking for – Khaled Oqab Jan 28 '19 at 09:07
  • Some related questions whose answers may be useful: https://math.stackexchange.com/questions/577781, https://math.stackexchange.com/questions/1211087, https://math.stackexchange.com/questions/2034864, and https://math.stackexchange.com/questions/2630257. A way to relate those answers to your formula would be to see how your formula relates to the other definitions of the determinant. – David K May 07 '19 at 13:04

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