Editing Invertible matrix (section)

=== Analytic solution ===
{{Main article|Cramer's rule}}
Writing the transpose of the [[matrix of cofactors]], known as an [[adjugate matrix]], may also be an efficient way to calculate the inverse of ''small'' matrices, but the [[Recursion|recursive]] method is inefficient for large matrices. To determine the inverse, we calculate a matrix of cofactors:
: <math>\mathbf{A}^{-1} = {1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}\mathbf{C}^\mathrm{T} =
  {1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}
  \begin{pmatrix}
    \mathbf{C}_{11} & \mathbf{C}_{21} & \cdots & \mathbf{C}_{n1} \\
    \mathbf{C}_{12} & \mathbf{C}_{22} & \cdots & \mathbf{C}_{n2} \\
             \vdots &          \vdots & \ddots &          \vdots \\
    \mathbf{C}_{1n} & \mathbf{C}_{2n} & \cdots & \mathbf{C}_{nn} \\
  \end{pmatrix}
</math>
so that
: <math>\left(\mathbf{A}^{-1}\right)_{ij} =
  {1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}\left(\mathbf{C}^{\mathrm{T}}\right)_{ij} =
  {1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}\left(\mathbf{C}_{ji}\right)
</math>
where {{math|{{abs|'''A'''}}}} is the [[determinant]] of {{math|'''A'''}}, {{math|'''C'''}} is the matrix of cofactors, and {{math|'''C'''<sup>T</sup>}} represents the matrix [[transpose]].

==== Inversion of 2 × 2 matrices ====
The ''cofactor equation'' listed above yields the following result for {{nowrap|2 × 2}} matrices. Inversion of these matrices can be done as follows:<ref>{{cite book
|title=Introduction to linear algebra
|edition=3rd
|first1=Gilbert
|last1=Strang
|publisher=SIAM
|year=2003
|isbn=978-0-9614088-9-3
|page=71
|url=https://books.google.com/books?id=Gv4pCVyoUVYC
}}, [https://books.google.com/books?id=Gv4pCVyoUVYC&pg=PA71 Chapter 2, page 71]
</ref>
: <math>\mathbf{A}^{-1} = \begin{bmatrix}
    a & b \\ c & d \\
  \end{bmatrix}^{-1} =
  \frac{1}{\det \mathbf{A}} \begin{bmatrix}
    \,\,\,d & \!\!-b \\ -c & \,a \\ 
  \end{bmatrix} =
  \frac{1}{ad - bc} \begin{bmatrix}
    \,\,\,d & \!\!-b \\ -c & \,a \\ 
  \end{bmatrix}.
</math>

This is possible because {{math|1/(''ad'' − ''bc'')}} is the [[reciprocal (mathematics)|reciprocal]] of the determinant of the matrix in question, and the same strategy could be used for other matrix sizes.

The Cayley–Hamilton method gives
: <math>\mathbf{A}^{-1} = \frac{1}{\det \mathbf{A}} \left[ \left( \operatorname{tr}\mathbf{A} \right) \mathbf{I} - \mathbf{A} \right] .</math>

==== Inversion of 3 × 3 matrices ====
A [[computationally efficient]] {{nowrap|3 × 3}} matrix inversion is given by
: <math>\mathbf{A}^{-1} = \begin{bmatrix}
    a & b & c\\ d & e & f \\ g & h & i\\
  \end{bmatrix}^{-1} =
  \frac{1}{\det(\mathbf{A})} \begin{bmatrix}
    \, A & \, B & \,C \\ \, D & \, E & \, F \\ \, G & \, H & \, I\\
  \end{bmatrix}^\mathrm{T} =
  \frac{1}{\det(\mathbf{A})} \begin{bmatrix}
     \, A & \, D & \,G \\ \, B & \, E & \,H \\ \, C & \,F & \, I\\
  \end{bmatrix}
</math>
(where the [[Scalar (mathematics)|scalar]] {{mvar|A}} is not to be confused with the matrix {{math|'''A'''}}).

If the determinant is non-zero, the matrix is invertible, with the entries of the intermediary matrix on the right side above given by
: <math>\begin{alignat}{6}
 A &={}&  (ei - fh), &\quad& D &={}& -(bi - ch), &\quad& G &={}&  (bf - ce), \\
 B &={}& -(di - fg), &\quad& E &={}&  (ai - cg), &\quad& H &={}& -(af - cd), \\
 C &={}&  (dh - eg), &\quad& F &={}& -(ah - bg), &\quad& I &={}&  (ae - bd). \\
\end{alignat}</math>

The determinant of {{math|'''A'''}} can be computed by applying the [[rule of Sarrus]] as follows:
: <math>\det(\mathbf{A}) = aA + bB + cC.</math>

The Cayley–Hamilton decomposition gives
: <math>\mathbf{A}^{-1} =
  \frac{1}{\det (\mathbf{A})}\left( \tfrac{1}{2}\left[ (\operatorname{tr}\mathbf{A})^{2} - \operatorname{tr}(\mathbf{A}^{2})\right] \mathbf{I} - \mathbf{A}\operatorname{tr}\mathbf{A} + \mathbf{A}^{2}\right). 
</math>

{{anchor|Inversion of 3×3 matrices based on vector products}}
The general {{nowrap|3 × 3}} inverse can be expressed concisely in terms of the [[cross product]] and [[triple product]]. If a matrix <math>\mathbf{A} = \begin{bmatrix} \mathbf{x}_0 & \mathbf{x}_1 & \mathbf{x}_2\end{bmatrix}</math> (consisting of three column vectors, <math>\mathbf{x}_0</math>, <math>\mathbf{x}_1</math>, and <math>\mathbf{x}_2</math>) is invertible, its inverse is given by 
: <math>\mathbf{A}^{-1} = \frac{1}{\det(\mathbf A)}\begin{bmatrix}
  {(\mathbf{x}_1\times\mathbf{x}_2)}^\mathrm{T} \\
  {(\mathbf{x}_2\times\mathbf{x}_0)}^\mathrm{T} \\
  {(\mathbf{x}_0\times\mathbf{x}_1)}^\mathrm{T}
\end{bmatrix}.</math>

The determinant of {{math|'''A'''}}, {{math|det('''A''')}}, is equal to the triple product of {{math|'''x'''{{sub|0}}}}, {{math|'''x'''{{sub|1}}}}, and {{math|'''x'''{{sub|2}}}}—the volume of the [[parallelepiped]] formed by the rows or columns: 
: <math>\det(\mathbf{A}) = \mathbf{x}_0\cdot(\mathbf{x}_1\times\mathbf{x}_2).</math>

The correctness of the formula can be checked by using cross- and triple-product properties and by noting that for groups, left and right inverses always coincide. Intuitively, because of the cross products, each row of {{math|'''A'''{{sup|–1}}}} is orthogonal to the non-corresponding two columns of {{math|'''A'''}} (causing the off-diagonal terms of <math>\mathbf{I} = \mathbf{A}^{-1}\mathbf{A}</math> be zero). Dividing by 
: <math>\det(\mathbf{A}) = \mathbf{x}_0\cdot(\mathbf{x}_1\times\mathbf{x}_2)</math>
causes the diagonal entries of {{math|1='''I''' = '''A'''{{sup|−1}}'''A'''}} to be unity. For example, the first diagonal is:
: <math>1 = \frac{1}{\mathbf{x_0}\cdot(\mathbf{x}_1\times\mathbf{x}_2)} \mathbf{x_0}\cdot(\mathbf{x}_1\times\mathbf{x}_2).</math>

==== Inversion of 4 × 4 matrices ====
With increasing dimension, expressions for the inverse of {{math|'''A'''}} get complicated. For {{math|1=''n'' = 4}}, the Cayley–Hamilton method leads to an expression that is still tractable:
:<math>\begin{align}
\mathbf{A}^{-1} =
  \frac{1}{\det(\mathbf{A})}\Bigl(
    &\tfrac{1}{6}\bigl( (\operatorname{tr}\mathbf{A})^{3} - 3\operatorname{tr}\mathbf{A}\operatorname{tr}(\mathbf{A}^{2}) + 2\operatorname{tr}(\mathbf{A}^{3})\bigr) \mathbf{I} \\[-3mu]
&\ \ \  - \tfrac{1}{2}\mathbf{A}\bigl((\operatorname{tr}\mathbf{A})^{2} - \operatorname{tr}(\mathbf{A}^{2})\bigr) + \mathbf{A}^{2}\operatorname{tr}\mathbf{A} -
    \mathbf{A}^{3}
  \Bigr).
\end{align}</math>