In a paper I'm studding the following is said:
If A is a square matrix then its matrix of cofactors is denoted by A*. The following identites are well known: A* A = A A* =det(A) I where I is the identity matrix. I'm using GNU Octave to test these assertions.
being: $$ A= \begin{bmatrix}2 & 1 & -2 \\ 1 & 1 & -2 \\ -1 & 0 & 1 \end{bmatrix} $$
I determinate A* by: $$ A^{*}=(A^{-1})^{T} \cdot |A| $$
$$ A^{*}= \begin{bmatrix}1 & 1 & 1 \\ -1 & 0 & -1 \\ 0 & 2 & 1 \end{bmatrix} $$
Finally $$ |A|= 1 $$
When I compute:
$$ A^{*} \cdot A = \begin{bmatrix}2 & 2 & -3 \\ -1 & -1 & 1 \\ 1 & 2 & -3 \end{bmatrix} $$
$$ A \cdot A^{*} = \begin{bmatrix}1 & -2 & 1 \\ 0 & -3 & -2 \\ -1 & 1 & 0 \end{bmatrix} $$
$$ |A| \cdot I = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
the last three matrices do not match as the paper said, can someone explain that I am doing wrong?
It turned out to be a problem of terminology of different bibliographic sources, in some books, the matrix of cofactors is known as the adjugate matrix and biceversa. In this paper when talking about the matrix of cofactors, reference was made to the adjugate matrix, that is, the transpose of the matrix of cofactors, when operating with the latter the proposed identities are correctly fulfilled.
The correct A* will be:
$$ A^{*}=\begin{bmatrix}1 & -1 & 0 \\1 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} $$