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.

  —

  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