Consider $A$ acting on a [complex vector space.](Complex%20vector%20spaces.md) $A,$ is invertible if and only if [$\det{(A)}\neq0$](Determinants.md) or equivalently if and only if $0$ is not an eigenvalue of $A.$ ([proof](Operator%20inverse.md#Proof%20that%20an%20operator%20is%20invertible%20if%20and%20only%20if%20it%20has%20a%20nonzero%20determinant))
%%Is this just for complex vector spaces? check this claim. Should be for some vector space over a field.%%
# Matrix inverse
## Properties of invertible matrices
1) If $A$ and $B$ are invertible then so is $AB$ ([proof](Operator%20inverse.md#Proof%20of%20property%201%20of%20invertible%20matrices)).
## Groups of Invertible matrices
Invertibility is a [group axiom](Group%20axioms.md) - which should hint at invertible matrices belonging in a [group](Groups.md). In fact, every invertible matrix is an element of the group [[GL(n;F)]].
# Finding the inverse
## Situations where the inverse is obvious
* For a [unitary operator](Unitary%20operators.md). $U$, its inverse is the [adjoint](Adjoint.md), $U^\dagger$.
* For an [orthogonal operator](Orthogonal%20operators.md), $O$, its inverse is given by its [transpose](Transpose%20of%20a%20linear%20map.md), $O^{T}$.
## Situations where the Inverse isn't obvious
### Inverse of a $2\times2$ matrix
For a $2 \times 2$ matrix that isn't unitary, and therefore where the inverse isn't immediately obvious,
we evaluate the inverse using the [determinant](Determinants.md):
$X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
$X^{-1}=\frac{1}{\det{X}}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$
### Inverse of a $3\times3$ matrix
No one should be asking you to calculate matrix inverses by hand for larger matrices. We have computers for that, but if you really need to know:
### Inverse of an $n\times n$ matrix
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# Proofs and Examples
## Proof of property 1 of invertible matrices
## Proof that an operator is invertible if and only if it has a nonzero determinant
#MathematicalFoundations/Algebra/AbstractAlgebra/LinearAlgebra/Operators/Matrices