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

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In summary, the problem to be solved is to prove that determinant(AB) is equal to the product of the determinants of A and B. The equations and strategies for solving this problem are discussed, including using an induction proof and knowledge about triangular matrices. Additionally, a method for proving the property using standard operations is mentioned, but it requires both A and B to adhere to the row-column rule.
  • #1
touch the sky
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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!
 
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  • #2
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.
 
  • #3
touch the sky said:

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

Enjoy :)
 
  • #4
Susanne217 said:
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 [tex]det(AB) = Det(A) \cdot Det(B)[/tex] can be proven using standard operations.
You remember that [tex]AB =[a_1, a_2,\ldots a_n] \cdot B[/tex]
This doesn't make any sense if you don't say what "[itex]a_1[/itex]", "[itex]a_2[/itex]", 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 [tex]Det(AB) = Det([a_1, a_2,\ldots a_n]) \cdot Det(B)[/tex] this fact works on both singular and non-singular cases.

Enjoy :)
 
  • #5
HallsofIvy said:
This doesn't make any sense if you don't say what "[itex]a_1[/itex]", "[itex]a_2[/itex]", 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 :)

[tex]a_1, a_2, \cdots a_n[/tex] 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 :)
 

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

1. What does it mean to prove that the determinant of AB is equal to the product of the determinants of A and B?

This means that the determinant of the matrix AB is equal to the determinant of matrix A multiplied by the determinant of matrix B. In other words, the product of the determinants of two matrices is equal to the determinant of the product of those matrices.

2. Why is it important to prove that the determinant of AB is equal to det(A)det(B)?

This is important because the determinant of a matrix is a key property that is used in many mathematical operations involving matrices. By proving that the determinant of AB is equal to the product of the determinants of A and B, we can simplify and solve more complex equations involving matrices.

3. How do you prove that the determinant of AB is equal to det(A)det(B)?

To prove this, we can use the properties of determinants and matrix multiplication. Specifically, we can use the fact that det(AB) = det(A)det(B) and that (AB)^-1 = B^-1A^-1. By using these properties and manipulating the equations, we can show that the determinant of AB is equal to det(A)det(B).

4. Can you provide an example of proving that the determinant of AB is equal to det(A)det(B)?

Yes, for example, let A = [2 0; 0 3] and B = [1 4; 3 2]. Then, det(A) = 6 and det(B) = -10. The product of A and B is AB = [2 8; 9 6], and the determinant of AB is -36. Using the equation det(AB) = det(A)det(B), we can see that -36 = 6 * (-10), which proves that the determinant of AB is equal to det(A)det(B).

5. Are there any special cases or exceptions to proving that the determinant of AB is equal to det(A)det(B)?

Yes, there are some special cases where this equation may not hold true. For example, if A and B are not square matrices or if one of the matrices is singular, the equation det(AB) = det(A)det(B) may not be valid. It is important to check the properties and conditions of the matrices before using this equation to prove the determinant of AB.

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