The determinant of a matrix can be interpreted as the signed volume of the unit square/cube/hypercube, etc.

Taking the latter as the definition of the determinant, I can understand why, in the $2\times2$ case, the determinant is computed in such a way, yet why is it that when dealing with a $3\times3$ matrix (or more generally, a $n \times n$ matrix) the determinant is computed the way it is?

Would appreciate any help!

Martin Sleziak
  • 50,316
  • 18
  • 169
  • 342
  • 3,725
  • 1
  • 11
  • 25
  • 4
    You might find [this post](https://math.stackexchange.com/q/668/81360) and [this post](https://math.stackexchange.com/q/3328630/81360) to be helpful – Ben Grossmann Sep 17 '19 at 11:51
  • If you have time and some basic knowledge on linear algebra, I strongly recommend http://www.math.umaine.edu/~weiss/books/linear_algebra_via_exterior_products.pdf which provides some great geometric intuitions behind determinant. Long story short, iterating over all possible permutations is important because you want determinant to change sigh (but only sigh) each time you swap two columns. For more details, check book I linked. – Janczar Knurek Sep 17 '19 at 13:01

0 Answers0