Let A be an n -by-n matrix. Then A is invertible if and only if det(A)≠0 .
27. Further properties of determinants
27. Further properties of determinants
Recall that (abcd) is invertible if and only if its determinant ad-bc is nonzero. We'll prove the analogue of this for n -by-n matrices.
We'll also prove the following:
det(AB)=det(A)det(B) .
Proof of the invertibility criterion
Recall that A is invertible if and only if its reduced echelon form (say Ared ) is the identity matrix. To get from A to Ared , we used some row operations. The effect of these on the determinant is:
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(type I) determinant unchanged,
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(type II) determinant rescaled by a nonzero factor,
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(type III) determinant switches sign.
Therefore det(A) differs from det(Ared) by a nonzero factor. In particular, det(A)≠0 if and only if det(Ared)≠0 .
If A is invertible, det(Ared)=det(I)=1 , so det(A)≠0 .
If A is not invertible, Ared has a row of zeros (Ared is a square matrix in reduced echelon form, so if every row is nonzero then the leading entries have to go along the diagonal, and you get Ared=I ; but we're assuming A is not invertible). Since Ared has a row of zeros, det(Ared)=0 , so det(A)=0 too.
Proof of multiplicativity of the determinant
We first prove det(AB)=det(A)det(B) in the case where A is an elementary matrix.
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If A=Eij(λ) then AB is obtained from B by Ri↦Ri+λRj , so det(AB)=det(B) (such a row operation doesn't change the determinant). We also have det(A)=1 , so we can verify the formula det(AB)=det(A)det(B) in this case just by calculating both sides.
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If A=Ei(λ) then AB is obtained from B by Ri↦λRi , so det(AB)=λdet(B) . We also have det(A)=λ , so again we can verify the formula just by computing both sides.
This shows that det(AB)=det(A)det(B) whenever A is an elementary matrix.
By induction, we can now show that det(AB)=det(A)det(B) whenever A is a product of elementary matrices, in other words whenever A is an invertible matrix.
If A is not invertible, then det(A)=0 (by the previous theorem) so to verify the formula in this case, we need to prove that det(AB)=0 . Suppose for a contradiction that det(AB)≠0 . Then AB is invertible. But if AB is invertible then B(AB)-1 is an inverse for A : A(B(AB)-1)=AB(AB)-1=I. This gives a contradiction, so we deduce that det(AB)=0 .
We have now checked the formula for all possible matrices A , which proves the theorem.