Find A^1 matrix 1 Attachment(s) How can I do this ? :( 
$\begin{pmatrix} 1 &1 &1 & &1 &0 &0\\ 0 &1 &1 & &0 &1 &0\\ 0 &0 &1 & &0 &0 &1 \end{pmatrix}$ $R1 \Rightarrow R1  R2$ $\begin{pmatrix} 1 &0 &0 & &1 &1 &0\\ 0 &1 &1 & &0 &1 &0\\ 0 &0 &1 & &0 &0 &1 \end{pmatrix}$ $R2 \Rightarrow R2  R3$ $\begin{pmatrix} 1 &0 &0 & &1 &1 &0\\ 0 &1 &0 & &0 &1 &1\\ 0 &0 &1 & &0 &0 &1 \end{pmatrix}$ $A^{1} =\begin{pmatrix} 1 &1 &0\\ 0 &1 &1\\ 0 &0 &1 \end{pmatrix}$ 
On a $\LaTeX$ note, the array environment is your friend. $\left(\begin{array}{@{}cccccc@{}} 1 &1 &1 &1 &0 &0\\ 0 &1 &1 &0 &1 &0\\ 0 &0 &1 &0 &0 &1 \end{array}\right)$ 
While technically Romsek's answer is correct, its a bit of overkill for a problem like this. You already have the matrix in upper triangular form which means it is trivial to solve $Ax = b$ for any $b$ you like. Choose $b$ to be a basis vector and back solve to determine each column of $A^{1}$ instantly. In general, this works anytime you have already obtained an LU decomposition. In this case, the $L$ is just the identity matrix. 
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1 Attachment(s) This is what I did to find A^1 And then? Is that the answer of all question? 
I guess Romsek found the A^1 more practically then I did.:) so how should I continuity. I learned how to find inverse of any 3x3 matrix so far now. 
after finding A^1 are we going to find x or b or both? 
1 Attachment(s) This is what I did. Did I continue correctly ? 
1 Attachment(s) Oops sorry for the orientation. I corrected. 
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