Proving Determinant(AB)=det(A)det(B)

[touch the sky]

Homework Statement

the problem is to prove that determinant(AB)=det(A)det(B)

Homework Equations

i don't think there is any equation. :(

The Attempt at a Solution

i can figure it out by taking arbitrary elements in rows and columns , but i was wondering if i can prove it in a more elegant way.
thanks a lot!

 

[Inferior89]

You can use a few different methods.
One way is to say that an invertible matrix is the product of elementary matrices, use some knowledge about determinants of elementary matrices (det(EA) = det(E)det(A) if E is elementary.) and do an induction proof.There is also another way where you need to know that:

You can turn a square matrix into a triangular matrix that doesn't change the determinant (up to a sign change).

The product of two triangular matrices is also triangular.

The determinant of a triangular matrix is the product of the diagonal elements.

 

[Susanne217]

Homework Statement

the problem is to prove that determinant(AB)=det(A)det(B)

Homework Equations

i don't think there is any equation. :(

The Attempt at a Solution

i can figure it out by taking arbitrary elements in rows and columns , but i was wondering if i can prove it in a more elegant way.
thanks a lot!

When I was a little girl I used to ask these types of question all the time, but discovered if you first read the paragraphs then you discover something wonderful like.

The property that $$det(AB) = Det(A) \cdot Det(B)$$ can be proven using standard operations.
You remember that $$AB =[a_1, a_2,\ldots a_n] \cdot B$$ if you then take the Det on both sides of the equality You will get $$Det(AB) = Det([a_1, a_2,\ldots a_n]) \cdot Det(B)$$ this fact works on both singular and non-singular cases.

Enjoy :)

 

[HallsofIvy]

When I was a little girl I used to ask these types of question all the time, but discovered if you first read the paragraphs then you discover something wonderful like.

The property that $$det(AB) = Det(A) \cdot Det(B)$$ can be proven using standard operations.
You remember that $$AB =[a_1, a_2,\ldots a_n] \cdot B$$

This doesn't make any sense if you don't say what "$a_1$", "$a_2$", etc. are to start with! Are they rows of A or columns of A? And what do you mean by that product?

if you then take the Det on both sides of the equality You will get $$Det(AB) = Det([a_1, a_2,\ldots a_n]) \cdot Det(B)$$ this fact works on both singular and non-singular cases.

Enjoy :)

 

[Susanne217]

This doesn't make any sense if you don't say what "$a_1$", "$a_2$", etc. are to start with! Are they rows of A or columns of A? And what do you mean by that product?

You are right :)

$$a_1, a_2, \cdots a_n$$ are row in the matrix A. I also forgot to mention for this be allowed then A and B must live up to the row column rule :)