Why do determinants have their particular form? I know that for a matrix , if then the matrix does not have an inverse, and hence the associated system of equations does not have a unique solution. However, why do the determinant formulas have the form they do? Why all the complicated co-factor expansions and alternating signs ? To sum it up: I know what determinants do, but its unclear to me why. Is there an intuitive explanation that can be attached to a co-factor expansion??..

Gaige Haynes

Gaige Haynes

Answered question

2022-09-05

Why do determinants have their particular form?
I know that for a matrix A, if det(A)=0 then the matrix does not have an inverse, and hence the associated system of equations does not have a unique solution. However, why do the determinant formulas have the form they do? Why all the complicated co-factor expansions and alternating signs ?
To sum it up: I know what determinants do, but its unclear to me why. Is there an intuitive explanation that can be attached to a co-factor expansion??..

Answer & Explanation

Raphael Singleton

Raphael Singleton

Beginner2022-09-06Added 19 answers

Two exercises that may give you the answer you need (no work, no gain):
1. Assume you have a square [ 0 , 1 ] × [ 0 , 1 ] in the ( x , y )-plane. Assume for some reason you need to change the variables you are using. The new variables you are using are now w = a x + b y and z = c x + d y, where a, b, c and d are numbers. What is the area of the original square under the new coordinate system, the ( w , z ) -plane?
2. A multi-linear mapping in R 2 (bilinear in this case) is a function, M : R 2 × R 2 R such that M ( a x + b x ^ , y ) = a M ( x , y ) + b M ( x ^ , y ) and M ( x , a y + b y ^ ) = a M ( x , y ) + b M ( x , y ^ ). The map is alternating if M ( x , y ) = M ( y , x ). These two properties are very useful. Exercise: Show that if has these properties then M ( x , y ) = k d e t ( x 1 y 1 x 2 y 2 )

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