Distributive Properties of the Determinate

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SUMMARY

The discussion centers on the distributive properties of the determinant in linear algebra, specifically addressing misconceptions about the expression "det(b + c) = det(b) + det(c)", which is incorrect. Participants clarify that the correct property is "det(AB) = det(A)det(B)", which relates to multiplication rather than addition. The definition of the determinant is provided, emphasizing its computation through permutations. Additionally, the conversation touches on the relationship between row operations and the determinant's properties.

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  • Familiarity with matrix multiplication and properties
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  • Basic understanding of permutations in mathematics
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Students and educators in linear algebra, mathematicians exploring determinant properties, and anyone seeking to clarify misconceptions about matrix operations and determinants.

John Creighto
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I don´t have my linear algebra books with me and I forget how the distributive property of the determinate is proven. Can someone point me to a good link_
 
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What distributive property are you talking about? The distributive property is a(b+ c)= ab+ ac. Where are you putting the determinant in that? If you are thinking "det(b+ c)= det(b)+ det(c)", that's simply not true.
 
HallsofIvy said:
What distributive property are you talking about? The distributive property is a(b+ c)= ab+ ac. Where are you putting the determinant in that? If you are thinking "det(b+ c)= det(b)+ det(c)", that's simply not true.

Distributive with respect to multiplication.
 
HallsofIvy said:
Do you mean "Det(AB)= det(A)det(B)"? That's now what I would call "distributive".

You might look at
https://www.physicsforums.com/showthread.php?t=94344

That's what they called it at mathworld.

The definition of the determinate is:

a=\sum_{j_1...j_n}(-1)^ka_{1j_1}a_{2j_2}...a_{nj_n}

where the sum is taken over all permutations of \{j_1...j_n\} and k=0 for even permutations and k=1 for odd permutations.
 
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Yes, they do call it that! If find that very peculiar.
 
HallsofIvy said:
Yes, they do call it that! If find that very peculiar.

I was thinking of this argument today. Given you can compute the determinate by row operations then it seems apparent given the associativity of matrices that first reducing one matrix to reduced row echelon form via row operations and then the other via row operations;

is equivalent to taking the composition of the two matrices which reduce the matricies for which the determinate is bing computed to row echelon form and then applying this transformation to the product of the matrix product for which the determinate is being computed.

However, I was hoping for a proof which directly applied the definition of the determinate.
 

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