Translation:
Given the matrices A=.... and B=....
a) Calculate $A^tA$ and $AA^t$, ($A^t$ is the transposed of the matrix $A$)
b) Calculate $A^{-1}$ (inverse matrix) and resolve the linear system $A$ ... = ....
c) Calculate $C^2$, where $C = ABA^t$
My question was about point b)
I thought that since
$$A \left( \begin{array}{c} x\\ y\\ z \end{array} \right) = \left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right) $$
It could be like:
$AX=B (A^{-1})AX=(A^{-1})B (I)X=(A^{-1})B X=(A^{-1})B $
But in general I don't know why it's saying that the inverse matrix of A is equal to the matrix A and (x y z) and that is equal to the inverse of matrix A and (1 1 1) so on so on...
I'm having trouble in seeing it so if you can help me it would be great.
EDIT: I can't really write the problem neither the images can be visualized because I don't have reputation. So you would have to copy the link and paste it on your browser.
You already know that $A^T.A=\operatorname{Id}$. But then $A^T=A^{-1}$ and therefore$$A.X=B\iff X=A^{-1}.B\iff X=A^T.B.$$