# Distributive Properties of the Determinate

#### John Creighto

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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#### HallsofIvy

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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.

#### John Creighto

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.

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#### John Creighto

Do you mean "Det(AB)= det(A)det(B)"? That's now what I would call "distributive".

You might look at
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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#### HallsofIvy

Homework Helper
Yes, they do call it that! If find that very peculiar.

#### John Creighto

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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