Matrices: Prove that A and B are both invertible, then AB is invertible and (AB)^-1 =B^-1 A^-1?

4 Answers

• Suppose that A and B are both invertible.

  Then, since A is invertible, then there exists a matrix A^-1 such that AA^-1 = I and A^-1A = I, where I is the identity matrix.

  Similarly, since B is invertible, then there exists a matrix B^-1 such that BB^-1 = I and B^-1B = I.

  To show that AB is invertible, all that one has to do is to demonstrate that it has an inverse; that is, we must exhibit a matrix C such that (AB)C = I, and C(AB) = I.

  Selecting B^-1A^-1 to be the matrix C works, because

  (AB)(B^-1A^-1) = A(BB^-1)A^-1 = A(I)A^-1 = (AI)A^-1 = AA^-1 = I; and

  (B^-1A^-1)AB = B^-1(A^-1A)B = B^-1(I)B = (B^-1I)B = B^-1B = I.

  (Here, I've used the facts that matrix multiplication is associative; that a matrix multiplied by its inverse yields the identity matrix; and that the identity matrix times any matrix is the other matrix.)

  So AB is invertible and B^-1A^-1 is the inverse of AB; in other words, B^-1A^-1 = (AB)^-1.

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  If AB is invertible, then yes, it will be true that both A and B are invertible.

  Suppose that AB is invertible. Then there exists a matrix C such that (AB)C = I and C(AB) = I.

  Since (AB)C = I and matrix multiplication is associative, then A(BC) = I. Thus, BC is an inverse for the matrix A. (To show that a square matrix has an inverse, it is enough to show that the multiplication works on one side only--if it does, then it will always work the other way also. That is to say, in this case, knowing that A(BC) = I, we can also conclude that (BC)A = I.)

  Also, since C(AB) = I, then (CA)B = I. Thus, CA is an inverse for the matrix B.

  Since A and B both have inverses, then both are invertible.

• The proof that if A and B are invertible, then AB is invertible can be done more elegantly if you know these two results:

  I. Det (AB) = (det (A))*(det(B)).

  II. A matrix B is invertible if and only if det(B) =/ 0.

  Proof. Suppose that both A and B are invertible. Then det(A) =/ 0 and det(B) =/ 0. Now by I, det(AB) =/ 0, so by II AB is invertible.

• Ditto for Awms. you're utilising a sledge hammer to rigidity a thumb tack. From AB = I, taking transposes we get B^T A^T = I^T = I. as a result, det(B^T)*det(A^T) = det(I) = a million, so the two det(A^T) and det(B^T) are non-0, so A^T and B^T are invertible. in addition to, B^T A^T = I says that B^T is the inverse of A^T; it particularly is, B^T = (A^T)^(-a million). What did we bypass away out?

• (AB)(B^-!A^-1)=

  A(BB^-1)A^-1=

  AIA^-1=AA^-1=I