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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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