Inversion calculation
Task number: 4561
Is the following procedure correct?
When programming the computation of the inverse matrix, we succeed only with Gaussian elimination, namely with performing the row transformations only in top-down manner, and matrix transpose: We modify the block matrix \((\boldsymbol A|\mathbf I)\) to \((\boldsymbol B|\mathbf C)\), where \(\boldsymbol B\) is upper triangular with ones on the diagonal. We then transpose both sides to \((\boldsymbol B^{\mathrm T}|\mathbf C^{\mathrm T})\), giving us a lower triangular matrix on the left side. Finally, we transform this block matrix again by Gaussian elimination to a block matrix, with the unit matrix on the left. On the right we then find the transpose of the matrix inverse of the matrix \(\boldsymbol A\), so we just transpose this to get \(\boldsymbol A^{-1}\). In other words, the second elimination gives the block matrix \((\mathbf I|(\boldsymbol A^{-1})^\mathrm{T})\).
Solution
Let us denote the resulting matrix \(\boldsymbol D\) and examine, how it depends on \(\boldsymbol A\).
Equivalent row transformations correspond to matrix multiplication from the left, and so the relation \((\boldsymbol A|\mathbf I)\sim(\boldsymbol B|\boldsymbol C)\) implies \(\boldsymbol{CA}=\boldsymbol B\). Similarly, from the relation \((\boldsymbol B^\mathrm{T}|\boldsymbol C^\mathrm{T})\sim\sim(\mathbf I|\boldsymbol D^\mathrm{T})\) implies \(\boldsymbol C^\mathrm{T}=\boldsymbol B^\mathrm{T}\boldsymbol D^\mathrm{T}\).
Thus \(\boldsymbol A=\boldsymbol C^{-1}\boldsymbol B\), and so \(\boldsymbol A^{-1}=(\boldsymbol C^{-1}\boldsymbol B)^{-1}=\boldsymbol B^{-1}\boldsymbol C\). However, the second equation \(\boldsymbol C^\mathrm{T}=\boldsymbol B^\mathrm{T}\boldsymbol D^\mathrm{T}\) implies \(\boldsymbol C=\boldsymbol{DB}\), and therefore \(\boldsymbol D=\boldsymbol C\boldsymbol B^{-1}\).
The matrix product is not commutative. To construct a counterexample, take the matrices \(\boldsymbol B^{-1}\) and \(\boldsymbol C\), whose product does not commute, while \(\boldsymbol B\) is upper triangular and \(\boldsymbol C\) is lower triangular. For example, for \(\boldsymbol B=\begin{pmatrix}1&2\\0&1\end{pmatrix}\) and \(\boldsymbol C=\begin{pmatrix}1&0\\3&1\end{pmatrix}\) we get \(\boldsymbol B^{- 1}=\begin{pmatrix}1&-2\\0&1\end{pmatrix}\) and \(\boldsymbol C^{-1}=\begin{pmatrix}1&0\\-3&1\end{pmatrix}\). From these matrices, we obtain the counterexample \(\boldsymbol A=\begin{pmatrix}1&2\\-3&-5\end{pmatrix}\), while \(\boldsymbol A^{-1}=\begin{pmatrix}-5&-2\\3&1\end{pmatrix}\ne\begin{pmatrix}1&-2\\3&-5\end{pmatrix}=\boldsymbol D\).
Answer
This simple and elegant procedure is, unfortunately, flawed.
