Math 2.0

Linear and multilinear algebra; matrix theory Proposition

Inverse of a product

Builds on Invertible matrix

Statement

Let RR be a commutative ring and let AA and BB be invertible n×nn \times n matrices over RR. Then ABAB is invertible, and (AB)−1=B−1A−1.(AB)^{-1} = B^{-1} A^{-1}.

Remarks

The proof is the computation (AB)(B−1A−1)=A(BB−1)A−1=AA−1=In(AB)(B^{-1} A^{-1}) = A (B B^{-1}) A^{-1} = A A^{-1} = I_n, and the same in the other order. By induction, (A1⋯Ak)−1=Ak−1⋯A1−1(A_1 \cdots A_k)^{-1} = A_k^{-1} \cdots A_1^{-1}. Thus the invertible n×nn \times n matrices form a group, the general linear group.

Reading path

To understand this page from the ground up, read these 6 pages in order. The foundations come first.

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Identity matrix, object, no evidence levelIdentity matrixobjectMatrix, definition, no evidence levelMatrixdefinitionMatrix multiplication, definition, no evidence levelMatrix multiplicationdefinitionSquare matrix, definition, no evidence levelSquare matrixdefinitionInvertible matrix, definition, no evidence levelInvertible matrixdefinitionInverse of a product, statement, no evidence levelInverse of a productstatement
Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Identity matrix object
  2. Matrix definition
  3. Matrix multiplication definition
  4. Square matrix definition
  5. Invertible matrix definition
  6. Inverse of a product statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Invertible_matrix, Properties
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UsesInvertible matrix

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