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If A is an elementary matrix and B is an arbitrary matrix of the same size then det(AB)=det(A)det(B)

Indeed, consider three cases:

Case 1. A is obtained from I by adding a row multiplied by a number to another row. In this case by the first theorem about elementary matrices the matrix AB is obtained from B by adding one row multiplied by a number to another row. Thus

det(AB)=det(B)


so we have:

det(AB)=det(A)det(B)


(both sides are equal to det(B)).

Case 2. A is obtained from I by multiplying a row by a number, say, k. In this case, again by the first theorem about determinants and the theorem about elementary matrices we have

det(A)=k, det(AB)=k det(B), so det(AB)=det(A)det(B).


Case 3. A is obtained from I by swapping two rows. In this case

det(A)=-1, det(AB)=-det(B), so again det(AB)=det(A)det(B).


The proof is complete.

Notice that this proof shows, in particular, that the determinant of any elementary matrix is not zero.
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A is invertible if and only if det(A) is not 0.

Suppose that A is invertible. Then by the second theorem about inverses A is a product of elementary matrices

A=E1E2...Ek


By the previous statement

det(A)=det(E1)det(E2)...det(Ek)


As we noticed before, none of the factors in this product is zero. Thus det(A) is not equal to zero.

Suppose now that A is not invertible. We need to prove that det(A)=0. Indeed, by the second theorem about inverses the reduced row echelon form of A is not I, so this reduced row echelon form (let us call it B) contains a zero row. Thus det(B)=0 by the second theorem about determinants. By the theorem about elementary matrices A is a product of some elementary matrices E1,...,Ek and B:

A=E1E2...EkB


Therefore by the previous property

det(A)=det(E1)det(E2)...det(Ek)det(B)=0


The proof if complete.
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det(AB)=det(A)det(B) for every A and B.

Suppose first that A is invertible. Then as before A is a product of elementary matrices

A=E1E2...Ek


so

AB=E1EE2...EkB


Then by the property a) of this theorem

det(AB)=det(E1)det(E2)...det(Ek)det(B)=det(A)det(B)


Now suppose that A is not invertible. Then by the property b) det(A)=0, so det(A)det(B)=0 and we need only prove that det(AB)=0. Since A is not invertible, by the second theorem about inverses the row echelon form C of the matrix A has a zero row. Therefore the matrix CB has a zero row (we noticed it before). Therefore det(CB)=0 (the second theorem about determinants). But A is equal to a product of elementary matrices times C:

A=E1E2...EkC


So AB is equal to a product of (the same) elementary matrices times CB:

AB=E1E2...EkCB


By the property a) of this theorem

det(AB)=det(E1)...det(Ek)det(CB)=0


The proof is complete.
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