# Is the inverse operation on matrices distributive?

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

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

Is it true for $1\times 1$ matrices?

• Thanks; no, it is not. I'll accept your answer in 12 minutes (which is apparently the required lag time). – tdenniston Oct 2 '13 at 19:50

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.

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 $$A=A^{-1}\pmatrix{1&1\\ 0&1},\quad B=B^{-1}\pmatrix{0&1\\ 1&0},\quad A+B=(A+B)^{-1}=\pmatrix{1&0\\ 1&1}.$$

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.