If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Linear algebra

### Course: Linear algebra>Unit 2

Lesson 7: Transpose of a matrix

# Determinant of transpose

Proof by induction that transposing a matrix does not change its determinant. Created by Sal Khan.

## Want to join the conversation?

• These lessons are starting to run together so correct me if I'm wrong, but this also holds true when A is m x n (m =/= n) as the det(A) of an mxn is just zero. Right?
• No determinant if rows don't equal columns.
• Why is the last value in the determinant not (-1)ⁿ⁺¹a_1m * det(A_1m)?
• Two things:

1.) You're right about the "a_1m" term--Sal accidently left it out.

2.) However, since Sal is expanding along the first row (and since their are m rows), the "-1" term should be raised to the "1+m" power (and not n+1). This ensures that if their are an even number of columns, the final minor is multiplied by (-1), and if their are an odd number of rows, the final minor is multiplied by 1. (Again, assuming that we--like Sal--expand along the first row of the matrix).

Thus, the last product should read as:
(-1)^(1+m)*(a_1m)*det(A_1m)
• At , the determinant of the transposed matrix is being calculated. Why would he be calculating it with a11, a12 (going down along the 1st column) rather than a11, a21 (going along the 1st row)?? Can you calculate a matrix like this?
• You can calculate the determinant by going down the rows, or going down the columns. It doesn't even have to be the first row, or first column, (as long as you put the negative sign for even rows or columns).
• In which video was it shown that you can do cofactor expansion down a row or column?
• Why can't I watch these videos??
(1 vote)
• If you can't watch them on khan academy go to YouTube and search under the same title.
Try refreshing page, making sure your system or iOS is updated or it might just be your Internet connection.
• Do you need to prove the nxn case?
(1 vote)
• That is what Sal does in this video. He accomplished this by induction on the size of the matrix. The induction works by first proving a base case, n=2 in this case. That was done first. The second step (and usually more difficult one) is proving that if we assume the theorem ( det A = det At ) is true in a particular case (n x n), then it must be the case that it's true in the next case ( n+1 x n+1 ). This second (inductive) step combined with the first (base) step, gives a proof of the theorem in all subsequent cases, sort of like a long line of dominoes set up for a chain reaction. Pushing over the first domino (base step), knocks it into the second, and the second is knocked into the third, and so on for the rest (inductive step). Hope that helps!