Inverses follow a few rules that come straight from undoing things in order. Undoing two steps means undoing the last one first, so the inverse of a product reverses the order of its factors.

You put on socks and then shoes, but you take off the shoes first. The inverse of a product is taken the same way round: last on, first off.

Every rule below is proved the same way: guess the inverse, multiply, and find the identity. An inverse is unique, so a matrix that gives $I$ from both sides is the inverse. For square matrices one side is even enough: if $BA=I$ then $AB=I$ as well. That fact is proved in the chapter on linear maps and used here.

• $(A^{-1})^{-1}=A$: undoing the undoing gives back the original map.
• $(AB)^{-1}=B^{-1}A^{-1}$ when $A$ and $B$ are invertible of the same order: the order reverses.
• $(A^{T})^{-1}=(A^{-1})^{T}$, and for .
• $(A^{k})^{-1}=(A^{-1})^{k}$, which is written .
• $AX=B$ is solved by $X=A^{-1}B$, and $XA=B$ by : the inverse goes on the side where stands.
• Sums have no rule: $(A+B)^{-1}$ is generally not $A^{-1}+B^{-1}$, and may not be invertible at all.

Multiply $AB$ by the candidate $B^{-1}A^{-1}$. Regroup so that $B$ meets its inverse in the middle, then $A$ meets its inverse, and the identity is left. The same happens in the other order, so the candidate is the inverse, and there is only one.

Products may be regrouped freely, because multiplication is associative.

$B$ meets its inverse in the middle.

The identity changes nothing, and then $A$ meets its inverse.

The other order regroups in the same way.

The candidate undoes $AB$ from both sides, and an inverse is unique.

$AB$ applies $B$ first and $A$ second. Undoing it undoes $A$ first, with $A^{-1}$, and then $B$, with $B^{-1}$. Written right to left, as products are, that is $B^{-1}A^{-1}$.

Matrices do not commute, so an equation must be multiplied by $A^{-1}$ on the same side on which $A$ stands, on both sides of the equation.

• $(AB)^{-1}=B^{-1}A^{-1}$: invert each factor and reverse the order.
• $(A^{-1})^{-1}=A$, $(A^{T})^{-1}=(A^{-1})^{T}$, $(cA)^{-1}=\tfrac1cA^{-1}$.
• $AX=B$ gives $X=A^{-1}B$; $XA=B$ gives .
• There is no rule for $(A+B)^{-1}$.

Great work! You now know how to:

• invert a product: $(AB)^{-1}=B^{-1}A^{-1}$;
• use the rules for powers, scalars and transposes;
• solve $AX=B$ and $XA=B$ with the inverse on the correct side;
• avoid inventing a rule for sums.