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For example, is the following true: $$(A + B)^{-1} = A^{-1} + B^{-1}$$ If $\det(A) \ne 0$, $\det(B) \ne 0$, and $\det(A + B) \ne 0$.

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    $\begingroup$ This is not true even for numbers. $\endgroup$ – njguliyev Oct 2 '13 at 19:49
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    $\begingroup$ No, and I bet you can come up with several very easy counter examples even with $\;2\times 2\;$ matrices $\endgroup$ – DonAntonio Oct 2 '13 at 19:49
  • $\begingroup$ Replace $A=I$ and $B=I$ and calculate inverses. $\endgroup$ – Widawensen Feb 22 '19 at 14:06
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Is it true for $1\times 1$ matrices?

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    $\begingroup$ Thanks; no, it is not. I'll accept your answer in 12 minutes (which is apparently the required lag time). $\endgroup$ – tdenniston Oct 2 '13 at 19:50
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Suppose that $(A+B)^{-1} = A^{-1}+B^{-1}$. Then,

$$\begin{align*} I &= (A+B)^{-1}(A+B) \\ &= (A^{-1}+B^{-1})(A+B) \\ &= A^{-1}A + A^{-1}B + B^{-1}A+B^{-1}B \\ &= 2I + A^{-1}B + B^{-1}A, \end{align*}$$ so $A^{-1}B+B^{-1}A = -I$ for all invertible $A$, $B$.

It should be easy to use this to construct a counterexample.

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No, it's not true in general. Just consider any field $\mathbb{F}$ of characteristic $0$ (e.g. $\mathbb{R}$ or $\mathbb{C}$) and any $n\times n$ invertible matrix $A$ over $\mathbb{F}$. Put $B=A$ and you'll see something's wrong.

With appropriate conditions, the statement can be true. For example, if the field is $GF(2)$ (that contains only $0$ and $1$, with $1+1=0$), then $\det(A)=1$ for all invertible $A$. It follows that $A^{-1}=\operatorname{adj}(A)$. Now, if $n$ happens to be equal to $2$, then adjugate matrices are additive. Hence $$ (A+B)^{-1}=\operatorname{adj}(A+B)=\operatorname{adj}(A)+\operatorname{adj}(B)=A^{-1}+B^{-1} $$ whenever $A,B$ and $A+B$ are invertible. For a concrete example, consider $$ A=A^{-1}\pmatrix{1&1\\ 0&1},\quad B=B^{-1}\pmatrix{0&1\\ 1&0},\quad A+B=(A+B)^{-1}=\pmatrix{1&0\\ 1&1}. $$

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For problems of the form $(A \pm \mathbf{uv}^T)\mathbf{x} = \mathbf{b}$, the Sherman-Morrison formula can be applied to address the inability of distribution of the matrix inverse.

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