Editing Invertible matrix (section)

== Derivative of the matrix inverse ==
Suppose that the invertible matrix '''A''' depends on a parameter ''t''. Then the derivative of the inverse of '''A''' with respect to ''t'' is given by<ref>{{cite book |first1=Jan R. |last1=Magnus |first2=Heinz |last2=Neudecker |title=Matrix Differential Calculus : with Applications in Statistics and Econometrics |location=New York |publisher=John Wiley & Sons |edition=Revised |year=1999 |pages=151–152 |isbn=0-471-98633-X }}</ref>
: <math> \frac{\mathrm{d}\mathbf{A}^{-1}}{\mathrm{d}t} = - \mathbf{A}^{-1} \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t} \mathbf{A}^{-1}. </math>

To derive the above expression for the derivative of the inverse of '''A''', one can differentiate the definition of the matrix inverse <math>\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}</math> and then solve for the inverse of '''A''':
: <math>
    \frac{\mathrm{d}(\mathbf{A}^{-1}\mathbf{A})}{\mathrm{d}t}
  = \frac{\mathrm{d}\mathbf{A}^{-1}}{\mathrm{d}t}\mathbf{A}
    + \mathbf{A}^{-1}\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}
  = \frac{\mathrm{d}\mathbf{I}}{\mathrm{d}t}
  = \mathbf{0}.
</math>

Subtracting <math>\mathbf{A}^{-1}\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}</math> from both sides of the above and multiplying on the right by <math>\mathbf{A}^{-1}</math> gives the correct expression for the derivative of the inverse:
: <math> \frac{\mathrm{d}\mathbf{A}^{-1}}{\mathrm{d}t} = - \mathbf{A}^{-1} \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t} \mathbf{A}^{-1}. </math>

Similarly, if <math>\varepsilon</math> is a small number then
: <math>\left(\mathbf{A} + \varepsilon\mathbf{X}\right)^{-1}
  = \mathbf{A}^{-1} - \varepsilon \mathbf{A}^{-1} \mathbf{X} \mathbf{A}^{-1} + \mathcal{O}(\varepsilon^2)\,.
</math>

More generally, if

: <math> 
\frac { \mathrm{d}f(\mathbf{A})}{ \mathrm{d}t} = \sum_i g_i (\mathbf{A}) \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}h_i (\mathbf{A}),
</math>

then,

: <math> f (\mathbf{A} + \varepsilon\mathbf{X}) = f (\mathbf{A}) + \varepsilon\sum_i g_i (\mathbf{A}) \mathbf{X} h_i (\mathbf{A}) + \mathcal{O}\left(\varepsilon^2\right).</math>

Given a positive integer <math>n</math>,

: <math>
\begin{align}
\frac{ \mathrm{d}\mathbf{A}^{n}}{ \mathrm{d}t} &=
   \sum_{i=1}^n \mathbf{A}^{i-1}\frac{ \mathrm{d}\mathbf{A}}{ \mathrm{d}t}\mathbf{A}^{n-i},\\
\frac{ \mathrm{d}\mathbf{A}^{-n}}{ \mathrm{d}t} &=
  -\sum_{i=1}^n \mathbf{A}^{-i}\frac{ \mathrm{d}\mathbf{A}}{ \mathrm{d}t}\mathbf{A}^{-(n+1-i)}.
\end{align}
</math>

Therefore,

: <math>
\begin{align}
(\mathbf{A} + \varepsilon \mathbf{X})^{n} &=
  \mathbf{A}^{n} + \varepsilon \sum_{i=1}^n \mathbf{A}^{i-1}\mathbf{X}\mathbf{A}^{n-i} + \mathcal{O}\left(\varepsilon^2\right),\\
(\mathbf{A} + \varepsilon \mathbf{X})^{-n} &=
  \mathbf{A}^{-n} - \varepsilon \sum_{i=1}^n \mathbf{A}^{-i}\mathbf{X}\mathbf{A}^{-(n+1-i)} + \mathcal{O}\left(\varepsilon^2\right).
\end{align}
</math>