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## Linear algebra

### Course: Linear algebra>Unit 2

Lesson 6: More determinant depth

# Duplicate row determinant

Determinant of a matrix with duplicate rows. Created by Sal Khan.

## Want to join the conversation?

• At , Sal mentioned that in the last couple of videos, we learnt that det(Sij)= -det(A). May I know where specifically I can find the proof of this result?
• We can do this theorem by induction.

1) This rule holds for all 2x2 matrices.
Let A be the matrix:
``a bc d``
and let S be the matrix:
``c da b``
(note that this is the only way to swap the rows of A)

Clearly, the determinant of A is ad-bc and the determinant of S is bc-ad, meaning det(S)=-det(A), proving the first part of the theorem.

2) Given that this rule holds for all (m-1)X(m-1) matrices, this rule holds for all mXm matrices.
Let's say we have a mXm matrix A such that Sij is as defined in this video.

A can be represented like this:
``[rows 1-->(i-1)]i[rows (i+1)-->(j-1)]j[rows (j+1)-->m]``
Sij can be represented like this:
``[rows 1-->(i-1)]j[rows (i+1)-->(j-1)]i[rows (j+1)-->m]``

Let's say we take the determinant of A and S about the row j for both matrices. For A, the matrix for each element corresponding to each element in row j will be (with the nth column removed):
``[rows 1-->(i-1)]i[rows (i+1)-->(j-1)][rows (j+1)-->m]``
For S, this matrix will be:
``[rows 1-->(i-1)][rows (i+1)-->(j-1)]i[rows (j+1)-->m]``

Let the former matrix be A[jn] and the latter matrix be Sij[jn]. As you can see, both of these matrices are (m-1)X(m-1) matrices and to transform A[jn] to Sij[jn], j-i-1 swaps were needed, so det(Sij[jn])=(-1)^(j-i-1)*det(A[jn]).

Also, remember that row j is now at row i in Sij. If j-i is even, then (-1)^(i+n)=(-1)^(j+n) for all natural n and there's no difference, but if j-i is odd, then (-1)^(i+n)=(-1)^(j+n)*-1 for all natural n, multiplying the coefficient of det(A) when calculating det(Sij) by -1.

Now let's review:
If j-i is even, then det(Sij)=(-1)^(j-i-1)*det(A). Since j-1 is even, j-i-1 is odd, so det(Sij)=-det(A).
If j-i is odd, then det(Sij)=(-1)^(j-i-i)*-1*det(A). Since j-1 is odd, j-i-1 is even so (-1)^(j-i-1) is 1 and thus multiplying it has no effect, meaning det(Sij)=-det(A).

Thus, this proves the second part of the theorem and since the first part of the theorem holds true, the theorem holds true for all mXm matrices such that m is natural and m>=2.

I hope this helps!
• At he says he addressed row swapping in a recent video. I remember nothing about that. What video was it?
• Explanation and proof of "row swapping" using the definition of a determinant:

"A square array of numbers bordered on the left and right by a vertical line and having a value equal to the algebraic sum of all possible products where the number of factors in each product is the same as the number of rows or columns, each factor in a given product is taken from a different row and column, and the sign of a product is positive or negative depending upon whether the number of permutations necessary to place the indices representing each factor's position in its row or column in the order of the natural numbers is odd or even." (from https://www.merriam-webster.com/dictionary/determinant)

For an nxn matrix the determinant has n! products of n terms, each with its number of permutations of subscripts (permutations of "11 22 33 44 ... nn" based on the row and column of each of its factors. Let's choose (without loss of generality) that the 1st subscript is in the same numerical order for all n! products. The first subscript of the factors in each product then goes through the rows in that order and the second subscript varies in n! permutations so that each factor comes from a different column.

If we exchange two rows, then there is an additional permutation of row subscripts in all n! products, without changing any of the second (column) subscripts. So all the signs of the products are reversed, which reverses the sign of their sum, the determinant.

In a completely similar way, exchanging two columns also reverses the sign of the determinant (Let the second (column) subscript be in the same numerical order for all the products and let the order of the row subscripts vary for each of the n! products).
• does this apply to duplicate columns as well?
• I have the same problem Bowen has, I can't figure out where Sal mentioned that in the last couple of videos, we learnt that det(Sij)= -det(A). Infact i'm somewhat a meticulous learner. Now I've wasted hours but that does not matters. What matters is knowing. Any help please? I wish to see where Khan said it. If it is a mistake, no problem I just want to know.
• It doesn't look like there's a video where he shows it really. Do any of the responses to Bowen's question help?
• Any mistakes here? Any comments?

Explanation and proof of "row swapping" using the definition of a determinant:

"A square array of numbers bordered on the left and right by a vertical line and having a value equal to the algebraic sum of all possible products where the number of factors in each product is the same as the number of rows or columns, each factor in a given product is taken from a different row and column, and the sign of a product is positive or negative depending upon whether the number of permutations necessary to place the indices representing each factor's position in its row or column in the order of the natural numbers is odd or even." (from https://www.merriam-webster.com/dictionary/determinant)

For an nxn matrix the determinant has n! products of n terms, each with its number of permutations of subscripts (permutations of "11 22 33 44 ... nn" based on the row and column of each of its factors. Let's choose (without loss of generality) that the 1st subscript is in the same numerical order for all n! products. The first subscript of the factors in each product then goes through the rows in that order and the second subscript varies in n! permutations so that each factor comes from a different column.

If we exchange two rows, then there is an additional permutation of row subscripts in all n! products, without changing any of the second (column) subscripts. So all the signs of the products are reversed, which reverses the sign of their sum, the determinant.

In a completely similar way, exchanging two columns also reverses the sign of the determinant (Let the second (column) subscript be in the same numerical order for all the products and let the order of the row subscripts vary for each of the n! products).