# derivative of inverse matrix

###### Theorem 1.

Suppose $A$ is a square matrix^{} depending on a real parameter $t$
taking values in an open set $I\mathrm{\subseteq}\mathrm{R}$. Further, suppose all
component^{} functions^{} in $A$ are differentiable^{}, and $A\mathit{}\mathrm{(}t\mathrm{)}$ is invertible^{}
for all $t$. Then, in $I$, we have

$$\frac{d{A}^{-1}}{dt}=-{A}^{-1}\frac{dA}{dt}{A}^{-1},$$ |

where $\frac{d}{d\mathit{}t}$ is the derivative.

###### Proof.

Suppose ${a}_{ij}(t)$ are the component functions for $A$, and ${a}^{jk}(t)$ are component functions for ${A}^{-1}(t)$. Then for each $t$ we have

$$\sum _{j=1}^{n}{a}_{ij}(t){a}^{jk}(t)={\delta}_{i}^{k}$$ |

where $n$ is the order of $A$, and
${\delta}_{i}^{k}$ is the Kronecker delta^{} symbol. Hence

$$\sum _{j=1}^{n}\frac{d{a}_{ij}}{dt}{a}^{jk}+{a}_{ij}\frac{d{a}^{jk}}{dt}=0,$$ |

that is,

$$\frac{dA}{dt}{A}^{-1}=-A\frac{d{A}^{-1}}{dt}$$ |

from which the claim follows. ∎

Title | derivative of inverse matrix |
---|---|

Canonical name | DerivativeOfInverseMatrix |

Date of creation | 2013-03-22 14:43:52 |

Last modified on | 2013-03-22 14:43:52 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 7 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 15-01 |