Finding inverse by elimination Find the inverse of the matrix $A$ below by elimination on [A I] By expanding the matrix into an alternating matrix.
$$
A=
\begin{bmatrix}
1 & -1 & 1 & -1 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
I hope the question meant this matrix below by "alternating matrix"
$$
A=
\begin{bmatrix}
1 & -1 & 1 & -1 & 1 & 0 & 0 & 0 \\
0 & 1 & -1 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & -1 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\
\end{bmatrix}
$$
My attempt keeps going like this, forming $U$ out of $A$ while the right hand side becomes $$L^{-1}$$
But since $U=A$ in this case $I=L^{-1}$
$$
[U L^{-1}]=
\begin{bmatrix}
1 & -1 & 1 & -1 & 1 & 0 & 0 & 0 \\
0 & 1 & -1 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & -1 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\
\end{bmatrix}
$$
making $U$ $I$ by multiplying sides by it's inverse.
$$
[I U^{-1}L^{-1}]=
\begin{bmatrix}
1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\
\end{bmatrix}
$$
So 
$$
A^{-1}=
\begin{bmatrix}
1 & 1 & 0 & 0\\
0 & 1 & 1 & 0 \\
0 & 0  & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
But the question asks us to "guess" the inverse? Is there any shortcut or something. I used the tradinitonal gaussian jordan elimination here to find the inverse but question says try to "guess" and "this is a special elemination for a special inverse" 
Can someone show me the way somehow?
 A: Look at the matrix $A$ and notice that adding the second row to the first yields $e_1=(1,0,0,0)$, adding the third to the second yields $e_2$ and adding the last to the third yield $e_3$, after these operations the matrix is reduced to the identity matrix. The inverse simply encodes these operations. It is straightforward to see what is in this case, but the reasoning is actually exactly the same as in your work. There is no way of guessing an inverse in general.
A: Guessing an inverse can sometimes be done if you know similar matrices and their inverses by heart. 
The matrices I mention in the comment look like this:
$${\bf D} = \left[\begin{array}{rrrrr}1&-1&0&0&0\\0&1&-1&0&0\\0&0&1&-1&0\\0&0&0&1&-1\\0&0&0&0&1\end{array}\right]$$
This matrix $\bf D$ approximates a differential operator as a discrete difference of neighbouring elements.
Now, it's inverse:
$${\bf D}^{-1} = \left[\begin{array}{rrrrr}1&1&1&1&1\\0&1&1&1&1\\0&0&1&1&1\\0&0&0&1&1\\0&0&0&0&1\end{array}\right]$$
Since scalar product with the 1-vector is a sum, these rows simply are the Riemann sums of function values supposedly stored in the vector you multiply with ( which approximates an integral ).
These are matrix implementations of very basic digital filters in signal processing. If you knew about these two beforehand, the matrix patterns would probably ring a bell or two.
