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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... enter image description here

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 $\tilde b=A_b^{-1}b$

Revolucion for Monica
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  • Where does the $-6$ come from in the fourth matrix? Generally, the more your product at the end looks like the identity matrix, the later your error was made. There are ways of determining this, but I'm no expert. – Mathematician 42 May 04 '16 at 10:58
  • Please, when performing row operations, write the steps that you took! It's so much easier with notation like $2R_1+R_2\rightarrow R_2$ so that it's easy to follow what you thought you were computing. – Michael Burr May 04 '16 at 11:12

1 Answers1

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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 $

Rebellos
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