I'm curious about is there any geometric property relative to negative value for determinant of matrix. det ( A ) &lt; 0I knew about some of determinant of matrix properties as following, but it seems to me that it is nothing relative to negative value of determinant of matrix det ( A B ) = det ( A ) d e t ( B ) ( M u l t i p l i c a t i v e ) det ( A ) = 0 ⟺ A is singular M 2 , 2 = [ a b c d ] | det ( M 2 × 2 ) | = | a d − b d | = volumn of parallelogram | det ( M n × n ) | = ∏ j = 1 n a i , j ( − 1 ) i + j det ( M i , j ) expansion of determinant alone the i t h row

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I&#039;m curious about is there any geometric property relative to negative value for determinant of matrix.  det      (    A    )    &amp;lt;  0I knew about some of determinant of matrix properties as following, but it seems to me that it is nothing relative to negative value of determinant of matrix  det      (    A    B    )    =  det      (    A    )    d  e  t      (    B    )    (  M  u  l  t  i  p  l  i  c  a  t  i  v  e  )  det      (    A    )    =      0      ⟺        A is singular        M          2      ,      2        =      [                            a                          b                                      c                          d                      ]    |  det  (            M              2        ×        2              )  |  =  |  a  d  −  b  d  |  =      volumn of parallelogram    |  det  (            M              n        ×        n              )    |  =      ∏          j      =      1        n        a          i      ,      j        (  −  1      )          i      +      j        det  (            M              i        ,        j              )        expansion of determinant alone the               i              t      h             row

aletantas1x · Accepted Answer

The geometric property of   ±  1 signed determinant matrices is the orientation of the vectors of the matrix in its given space(for e.g             R              n      ) with respect to a fixed orientation of the basis vectors. If the vectors of your matrix   A has the same orientation as the basis then   s  i  g  n  (  d  e  t  (  A  )  )  =  +  1 otherwise its   −  1.For an example, take the matrix   A  =  (      e          2        ,      e          1        ,      e          3        )  =  (  (  0  ,  1  ,  0  )  ,  (  1  ,  0  ,  0  )  ,  (  0  ,  0  ,  1  )  ) in             R              3      . Then think of the orientation of the standard basis axes,   (      e          1        ,      e          2        ,      e          3        ), in             R              3       to be       c    l    o    c    k    w    i    s    e   such that if you drew an equilateral triangle in the             R              3       plane intersecting   (      e          1        ,      e          2        ,      e          3        ), then to get from       e          1       to       e          2       to       e          3       you would have to travel around the triangle in a clockwise rotation. If you perform the same triangle exercise with the vector   (      e          2        ,      e          1        ,      e          3        ) you would see that you move   c  o  u  n  t  e  r  c  l  o  c  k  w  i  s  e around the triangle. Thus, the matrix   A  =  (      e          2        ,      e          1        ,      e          3        ) has the opposite orientation as the standard basis and   s  i  g  n  (  d  e  t  (  A  )  )  =  −  1.This also explains why when you swap two columns/rows of a matrix the sign of the determinant changes with each swap. Unfortunately though, this geometric idea is a bit more difficult to understand in higher dimensions.

I'm curious about is there any geometric property relative to negative value for determinant of matr

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