Matrix Inversion

The iverse of the matrix A is to solve the equation $$\textbf{AX} = \textbf{I}, $$ where \(I\) is the \(n\times n\) identity matrix. The solution \(\textbf{X}\), also of size \(n\times n\), which will be the inverse of \(\textbf{A}\). The proof: after the mutiplication of both size of the equation is done by \(\textbf{A}^{-1}\) the following equation is obtained \(\textbf{A}^{-1}\textbf{A} \textbf{X} = \textbf{A}^{-1} \textbf{I}\),which in the end reduces to \(\textbf{X} = \textbf{A}^{-1}\).

The inversion of large matrices should be avoided dze to its high computational cost. The inversion of \(\textbf{A}\) is equivalent to solving \(\textbf{Ax}_i = \textbf{b}_i\) with \(i = 1,2,...,n,\) where \(\textbf{b}_i\) is the ith column of matrix \(\textbf{b}_i\). For example if LU decomposion method is employed in the solution, te solution phase whihc consists of forward and backward substitution must be repeated n times, once for each \(\textbf{b}_i\). Since the computational costs are proportional to \(n^3\) for the decomposition phase and \(n^2\) for each vector of the solution phase, the cost of the inversion is considerably more expensive than the solution of \(\textbf{Ax} = \textbf{b}\).

Formula for matrix inversion of second order matrix

For matrix \(\textbf{A}=\begin{bmatrix}a & b \\ c & d\end{bmatrix}\) the inverse can be determined using following expression $$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$

\(a\) =		\(b\) =
\(c\) =		\(d\) =
\(A^{-1}\) =