I don't know why but I'm really really weak in inverting matrices since years... I always do some mistakes. I'm asking you how could I cope with that problem and be able to invert matrix easily in the future.
For instance I tried to calculate the invert matrix $A_B^{-1}$
$$A_B=\begin{pmatrix} 1 & 2 & 3 & 0\\ 1 & 0 & 1 & 0\\ 1 & 1 & -3 & 1\\ 1 & 0 & 3 & 0 \end{pmatrix}$$
I started
$$\begin{pmatrix} 1 & 2 & 3 & 0 & | 1 & 0 & 0 & 0\\ 1 & 0 & 1 & 0& | 0 & 1 & 0 & 0\\ 1 & 1 & -3 & 1& | 0& 0 & 1 & 0\\ 1 & 0 & 3 & 0& | 0 & 0 & 0 & 1\\ \end{pmatrix}$$
$$\begin{pmatrix} 1 & 0 & 3 & 0& | 0 & 0 & 0 & 1\\ 1 & 2 & 3 & 0 & | 1 & 0 & 0 & 0\\ 1 & 0 & 1 & 0& | 0 & 1 & 0 & 0\\ 1 & 1 & -3 & 1& | 0& 0 & 1 & 0\\ \end{pmatrix}$$
$$\begin{pmatrix} 1 & 0 & 3 & 0& | 0 & 0 & 0 & 1\\ 0 & 2 & 0 & 0 & | 1 & 0 & 0 & -1\\ 0 & 0 & -2 & 0& | 0 & 1 & 0 & -1\\ 0 & 1 & -3 & 1& | 0& 0 & 1 & -1\\ \end{pmatrix}$$
$$\begin{pmatrix} 1 & 0 & 3 & 0& | 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & | 1/2 & 0 & 0 & -1/2\\ 0 & 0 & 1 & 0& | 0 & -1/2 & 0 & 1/2\\ 0 & 1 & -6 & 1& | 0& 0 & 1 & -1\\ \end{pmatrix}$$
$$\begin{pmatrix} 1 & 0 & 3 & 0& | 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & | 1/2 & 0 & 0 & -1/2\\ 0 & 0 & 1 & 0& | 0 & -1/2 & 0 & 1/2\\ 0 & 0 & -6 & 1& | -1/2& 0 & 1 & -1/2\\ \end{pmatrix}$$
$$\begin{pmatrix} 1 & 0 & 0 & 0& | 0 & -3/2 & 0 & 5/2\\ 0 & 1 & 0 & 0 & | 1/2 & 0 & 0 & -1/2\\ 0 & 0 & 1 & 0& | 0 & -1/2 & 0 & 1/2\\ 0 & 0 & 0 & 1& | -1/2& -3 & 1 & -5/2\\ \end{pmatrix}$$
Then I did $A_B^{-1}A_B$ for the first time in order to know if I have $I$
\begin{align*}A_B^{-1}A_B&=\begin{pmatrix} 0 & -3/2 & 0 & 5/2\\ 1/2 & 0 & 0 & -1/2\\ 0 & -1/2 & 0 & 1/2\\ -1/2& -3 & 1 & -5/2\\ \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 0\\ 1 & 0 & 1 & 0\\ 1 & 1 & -3 & 1\\ 1 & 0 & 3 & 0 \end{pmatrix}\\&= \begin{pmatrix} 1 & 0 & -3/2+15/2 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & -12/2-15/2 & 1\\ \end{pmatrix}\\ &\neq I \end{align*}
I therfore know that I'm wrong on the invert. From the result of $A_B^{-1}A_B$ How should I know were I was wrong? Is there a simpler method?
This comes from a linear programing problem...
I have to show that the basis $B=\{1,2,4,7\}$ is feasible by giving $\tilde b$ of the associate dictionnary:
$$x_B=\tilde b-\tilde A_{\bar B}x_{\bar B},\bar B =\{3,5,6,8\}$$
with $\tilde b=A_b^{-1}b$
You can always use the adjugate if you know it's mathematical definition. Then $A^{-1} $ is written as :
$A^{-1} = \frac{1}{det(A)}adj(A) $ where : $adj(A)$ is the transpose of the cofactor matrix C of A. The cofactor matrix of $A$ is the n×n matrix $C$ whose $(i, j)$ entry is the $(i, j)$ cofactor of $A$. Otherwise your operation is correct in terms of logic (your mistake is at the execution), because : $ AA^{-1} = A^{-1}A = I $