Det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd Why?

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In summary, the determinant of a matrix A is equal to the determinant of its transpose and the negative of A. This can be shown using the definition of the determinant and multiplying a single row by a scalar. However, this only holds true when the size of the matrix is even. For odd-sized matrices, the determinant will always be 0. This is relevant for skew symmetric matrices, where the determinant will always be 0.
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det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd
I understand all of this except for the last statement: =(-1)^n*det(A).
Why is this? Does it have to do with the cofactor expansion, and the "checkerboard" (as my professor puts it) of signs? I'm just not quite sure why that has to be. A quick explanation would be more than adequate. Thank you very much for your time. I didn't think that my post fit the rest of the template so I deleted it.
 
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It follows from the fact that: if B is formed from multiplying a single row of A by scalar x, then det(B) = x(det(A)). This can be shown easily using the definition of the determinant - you know there is some bij = xaij (where i is the row you multiplied to), so you can factor out the x. The desired result comes from multiplying all rows of A by x.

Also, note that det(A) = det(AT) is true but det(A) = det(-A) only if the size of n is even, and your statement: "det(A) = 0 if n when n is odd" doesn't make sense. det(A)=0 tells you about the invertibility of A.
 
  • #3
VeeEight said:
It follows from the fact that: if B is formed from multiplying a single row of A by scalar x, then det(B) = x(det(A)). This can be shown easily using the definition of the determinant - you know there is some bij = xaij (where i is the row you multiplied to), so you can factor out the x. The desired result comes from multiplying all rows of A by x.

Also, note that det(A) = det(AT) is true but det(A) = det(-A) only if the size of n is even, and your statement: "det(A) = 0 if n when n is odd" doesn't make sense. det(A)=0 tells you about the invertibility of A.

Wow I'm an idiot. I forgot to mention that I was talking about a skew symmetric matrix. So I guess it does make sense. Thank you for your help.
 
  • #4
Okay, I thought you were talking about something specific but thought I'd throw in that last bit anyway to be safe. Cheers.
 

1. What is the determinant of a matrix A?

The determinant of a matrix A is a scalar value that represents the scaling factor of the matrix transformation. It is denoted as det(A) or |A|.

2. What is the relationship between the determinant of a matrix and its transpose?

The determinant of a matrix A is equal to the determinant of its transpose, det(A) = det(AT). This means that the determinant of a matrix is unaffected by transposition.

3. How is the determinant of a negative matrix related to the determinant of the original matrix?

The determinant of a negative matrix -A is equal to the negative determinant of the original matrix, det(-A) = -det(A). This means that multiplying a matrix by -1 will result in the opposite sign of its determinant.

4. Why is the determinant of a matrix A raised to the power of its dimensions equal to the determinant of the matrix A?

This is true because each time a matrix is transposed, its dimensions are also switched. Therefore, raising the determinant of a matrix A to the power of its dimensions is the same as taking the determinant of the transposed matrix AT, which is equal to det(A).

5. Why is the determinant of a matrix equal to 0 when the dimensions of the matrix are odd?

This is due to the fact that when a matrix is transposed, its dimensions are switched and multiplying the determinant by -1 results in the opposite sign. Therefore, when the dimensions are odd, the determinant of the matrix AT will be equal to -det(A), which will cancel out the original determinant of det(A), resulting in a determinant of 0.

Suggested for: Det(A) = det(AT) = det(−A) = (−1)^n*det(A). Hence det(A) = 0 when n is odd Why?

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