Refresher - Inverse

Refresher

In linear algebra an n-by-n (square) matrix A is called invertible (some authors use nonsingular or nondegenerate) if there exists an n-by-n matrix B such that

${\displaystyle AB=BA=I_n}$

where ${\displaystyle I_n}$ denotes the n-by-n identitymatrix and the multiplication used is ordinary matrixmultiplication. If this is the case, then the matrix B is uniquely determined by A and is called the inverse of A, denoted by ${\displaystyle A^{-}1}$. It follows from the theory of matrices that if

${\displaystyle AB=I}$

A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if and only if its determinant is 0.

If ${\displaystyle A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$, the inverse is denoted by ${\displaystyle A^{-1}}$. In order to be inverses, ${\displaystyle A?A^{-1}=I}$, where I is the identity matrix.

In general, the inverse of the 2×2 matrix ${\displaystyle A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$ is given by:

${\displaystyle A^{-1}}$= ${\displaystyle \frac{1}{detA}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]}$ , where ${\displaystyle detA=ad-bc}$.

Note: This only works for 2 × 2 matrices.

Inversion of 3×3 matrices

A computationally efficient 3x3 matrix inversion is given by

${\displaystyle A^{-1}={\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& k\end{array}\right]}^{-1}=\frac{1}{det\left(A\right)}{\left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& K\end{array}\right]}^T=\frac{1}{det\left(A\right)}\left[\begin{array}{ccc}A& D& G\\ B& F& K\\ C& F& K\end{array}\right]}$

where the determinant of A can be computed by applying the rule of Sarrus as follows:

${\displaystyle det\left(A\right)=a(ek-fh)+b\left(fg-kd\right)+c(dh-eg)}$

If the determinant is non-zero, the matrix is invertible, with the elements of the above matrix on the right side given by

${\displaystyle A=(ek-fh)}$ ${\displaystyle D=(ch-bk)}$ ${\displaystyle G=\left(bf-ce\right)}$
${\displaystyle B=\left(fg-dk\right)}$ ${\displaystyle E=(ak-cg)}$ ${\displaystyle H=(cd-af)}$
${\displaystyle C=(dh-eg)}$ ${\displaystyle F=(db-ah)}$ ${\displaystyle K=(ae,bd)}$