= Invertibility =

'''Invertibility''' is a property of square matrices. If a matrix is invertible, there is an inverse matrix that it can be multiplied by to produce the [[LinearAlgebra/SpecialMatrices#Identity_Matrix|identity matrix]]. The calculation of an inverse matrix is '''inversion'''.

<<TableOfContents>>

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== Definition ==

A matrix '''''A''''' is invertible if there is a matrix '''''A'''^-1^'' which satisfies '''''AA'''^-1^ = '''A'''^-1^'''A''' = '''I'''''.

If a matrix cannot be inverted, it is '''singular''' and '''degenerate''' and '''non-invertible'''.

Only square matrices can be invertible. However, a non-square matrix can separably have distinct '''left inverse''' and '''right inverse''' matrices. Generally, if ''m < n'', then a matrix with shape ''m'' by ''n'' and rank of ''m'' can have a right inverse; a matrix with shape ''n'' by ''m'' and rank of ''m'' can have a left inverse.



=== Properties ===

By definition, '''''AA'''^-1^ = '''A'''^-1^'''A''' = '''I'''''.

An invertible matrix has only one vector in the [[LinearAlgebra/NullSpace|null space]]: the zero vector. If any basis vector of a matrix is a linear transformation of another, then the matrix does not have [[LinearAlgebra/Basis|basis]] and must be non-invertible.

For [[LinearAlgebra/Orthogonality#Matrices|orthogonal matrices]] (such as [[LinearAlgebra/SpecialMatrices#Permutation_Matrices|permutation matrices]]), the inverse is also the [[LinearAlgebra/Transposition|transpose]]: '''''Q'''^-1^ = '''Q'''^T^''.

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== Test with Determinant ==

The [[LinearAlgebra/Determinant|determinant]] is the most common test for invertibility. If ''|'''A'''| != 0'', then '''''A''''' is invertible. If ''|'''A'''| = 0'', then '''''A''''' is non-invertible.

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== Calculation with Elimination ==

Because '''''AA'''^-1^ = '''I''''', applying [[LinearAlgebra/Elimination|elimination]] and [[LinearAlgebra/Elimination#Reduced_Row_Echelon_Form|backwards elimination]] on '''''A''''' augmented with an [[LinearAlgebra/SpecialMatrices#Identity_Matrix|identity matrix]] ('''''I''''') will create '''''A'''^-1^'' in the augmentation.

{{{
┌            ┐
│ [1] 3 │ 1 0│
│  2  7 │ 0 1│
└            ┘
┌               ┐
│ [1]  3  │  1 0│
│  0  [1] │ -2 1│
└               ┘
┌             ┐
│ 1  3  │  1 0│
│ 0 [1] │ -2 1│
└             ┘
┌                ┐
│ [1]  0  │  7 -3│
│  0  [1] │ -2  1│
└                ┘
}}}

'''''A'''^-1^'' is:

{{{
┌      ┐
│  7 -3│
│ -2  1│
└      ┘
}}}

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== Calculation with Determinants and Cofactor Matrices ==

Given the [[LinearAlgebra/Determinant|determinant]] of '''''A''''', it can also be simple to compute '''''A'''^-1^'' as ''(1/|'''A'''|)'''C'''^T^''. '''''C''''' is the cofactor matrix, where ''c,,i j,,'' is the cofactor of ''a,,i j,,''.

For example, given a 2 x 2 '''''A''''' like:

{{{
┌    ┐
│ a b│
│ c d│
└    ┘
}}}

The cofactor matrix '''''C''''' is:

{{{
┌      ┐
│  d -c│
│ -b  a│
└      ┘
}}}

But this must be transposed to '''''C'''^T^'':

{{{
┌      ┐
│  d -b│
│ -c  a│
└      ┘
}}}

And then '''''A'''^-1'' is:

{{{
┌                               ┐
│  (1/det A) * d  (1/det A) * -b│
│ (1/det A) * -c   (1/det A) * a│
└                               ┘
}}}

The above example fits into this formula. The [[LinearAlgebra/Elimination|elimination]] and [[LinearAlgebra/Elimination#Reduced_Row_Echelon_Form|backwards elimination]] prove that the determinant of that '''''A''''' is 1. The more fundamental formula ''ad - bc'' expands to ''1 * 7 - 2 * 3'' which also reveals a determinant of 1. As such, ''(1/|'''A'''|)'' is trivially 1. So simply plug the given (''a'', ''b'', ''c'', ''d'') into the transposed cofactor matrix to find the inverse.



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