Editing Invertible matrix (section)

==== 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>