Math 2.0

statement proposition

Inverse of a product

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.

Explanations

L1 human

To undo two steps, we undo the last step first. If we first apply BB and then AA, then we must first undo AA and then undo BB. So the inverse of ABAB is B−1A−1B^{-1} A^{-1}, with the order reversed.

What this node needs for W1
  • an inventory row: the row of the inventory that the node accounts for (0 of 1)

The writing levels say how complete the text of a node is. They do not measure its truth.

What this needs

To read this node, read these 6 nodes in this order. The foundations come first, and the list ends with this node.

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 node uses the nodes above it. The colour of the bar on a box shows the evidence level, and the box also gives it as text. The list below has the same nodes.
  • no evidence level
  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

Sources

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

What links here

No node links to this node yet.