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### Course: Linear algebra>Unit 2

Lesson 6: More determinant depth

# Determinant after row operations

What happens to the determinant when we perform a row operation. Created by Sal Khan.

## Want to join the conversation?

• is this true even if you reduce it all the way to Reduce row echelon form?
• No. In reduced row echelon form you usually have to divide rows in order to get those leading 1's. When you divide a row the determinant will also be divided and thus will change.
• Couldn't we just immediately note that det([r_1; ...; r_i; ...; -cr_i; ... r_n]) = 0 because rows i and j are linearly dependent?
• Yes. If one row is a multiple of another, you can use this to zero out one of the rows, and if you you can expand along that 0 row to show clearly that the determinant is 0.
• Is this still true if you are altering the first row? For example doing a row operation such as row 1 = row 1 - row 2
• Yes. Here j=1, i=2, and c=1. it was drawn with i < j, but this was a simplification where we lose no generality, allowing i,j to take on values for any two distinct rows. Hope that helps!
• So, can I say that duplicate rows in a matrix result in a 0 determinant b/c that means that the whole matrix is squished onto a single line? If so, what happens to the other vectors that are not duplicates?
• Q. Let A be a square matrix of order 3 x 3, then k|A| is equal to
(A) k|A|
(B) k^2|A|
(C) k^3|A|
(D) 3k |A|

The correct answer is (C), can anyone expain how?
(1 vote)
• Do you mean |kA|?
Because k|A| is equal to k|A|.

To compute |kA|, you need to know that everytime you scale a row of a matrix, it scales the determinant. There are 3 rows in A, so kA is A with 3 rows scaled by k, which multiplies the determinant of A by k^3.

In general if A is n x n, then |kA|=k^n |A|.
• Does that mean I can bring any matrix in rref and then find the determinant of that matrix, which is easier because it has a lot of zeros?
(1 vote)
• Yes, that would make the determinant easier to compute, at the cost of having to compute the row operations first.
(1 vote)
• How to prove that:-

| a^2 +1 ab ac|
|ab b^2+1 bc| = 1+a^2+b^2+c^2
|ca cb c^2+1|
(1 vote)