Find the inverse of a $4\times4$ matrix My matrix looks like this:
$$\left(\begin{array}{rrrr}
    1&  1 & 1 & 1\\
    1& -1 & 1 & 0\\
    1&  1 & 0 & 0\\
    1&  0 & 0 & 0 
\end{array}\right)$$
The right lower half are all zeros.
Is there a quick way to find an inverse of this matrix?
I have the solution, but I'm unable to find the algorithm to get the inverse.
 A: Another way.  Let your matrix be $A$.  Then $A = J B$ where $$J = \pmatrix{0&0&0&1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0\cr}, \ B = JA = \pmatrix{1 & 0 & 0 & 0\cr
1 & 1 & 0 & 0\cr 1 & -1 & 1 & 0\cr 1 & 1 & 1 & 1\cr}$$ since multiplying by $J$ on the left flips the matrix vertically (interchanging first and fourth rows and second and third rows). 
Now $B$ is a lower triangular matrix.  We can write it as $I + N$ where the nonzero entries of $N$ are all below the main diagonal.  Thus $N^4 = 0$ and 
$$ B^{-1} = I - N + N^2 - N^3 =\pmatrix{ 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr} - \pmatrix{ 0 & 0 & 0 & 0 \cr
1 & 0 & 0 & 0 \cr 1 & -1 & 0 & 0 \cr 1 & 1 & 1 & 0 \cr}
+ \pmatrix{ 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & 0 \cr -1 & 0 & 0 & 0 \cr 2 & -1 & 0 & 0 \cr}
- \pmatrix{ 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr -1 & 0 & 0 & 0 \cr}
= \pmatrix{ 1 & 0 & 0 & 0 \cr
-1 & 1 & 0 & 0 \cr -2 & 1 & 1 & 0 \cr 2 & -2 & -1 & 1 \cr}
$$
We'll then have $$A^{-1} = B^{-1} J^{-1} = B^{-1} J = \pmatrix{ 0 & 0 & 0 & 1 \cr
0 & 0 & 1 & -1 \cr 0 & 1 & 1 & -2 \cr 1 & -1 & -2 & 2 \cr}
$$ where multiplying by $J$ on the right flips the matrix horizontally (interchanging first and fourth columns and second and third columns).  
A: You can also do it by inspection. Let $X$ be the inverse, and $\overline{x_i}$ the i-th row.
Then
$$
\left(\begin{array}{rrrr}
1 & 1 & 1 & 1\\
1 & -1 & 1 & 0\\
1 & 1 & 0 & 0\\
1 & 0 & 0 & 0
\end{array}\right)
X
= I
$$
So, considering first the last row, we get
$$ (1 \;  \phantom{-}0 \; \phantom{-}0 \; \phantom{-}0) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4}  \end{array}\right) = \overline{x_1} = (0 \; \phantom{-}0 \; \phantom{-}0 \; \phantom{-}1) $$
Then, from row 3 we get
$$(1 \;  \phantom{-}1 \; \phantom{-}0 \; \phantom{-}0) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4} \end{array}\right) 
= \overline{x_1} + \overline{x_2} =  (0 \; \phantom{-}0 \; \phantom{-}1 \; \phantom{-}0)  \Rightarrow \overline{x_2} = (0 \; \phantom{-}0 \; \phantom{-}1 \; -1)
$$
and 
$$(1 \;  -1 \; \phantom{-}1 \; \phantom{-}0) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4} \end{array}\right) 
= \overline{x_1} - \overline{x_2} + \overline{x_3} =  (0 \; \phantom{-}1 \; \phantom{-}0 \; \phantom{-}0)  \Rightarrow \overline{x_3} = (0 \; \phantom{-}1 \; \phantom{-}1 \; -2)
$$
$$(1 \;  \phantom{-}1 \; \phantom{-}1 \; \phantom{-}1) \left(\begin{array}{r} \overline{x_1} \\ \overline{x_2} \\ \overline{x_3} \\ \overline{x_4}  \end{array}\right) 
= \overline{x_1} + \overline{x_2} + \overline{x_3} + \overline{x_4} =  (1 \; \phantom{-}0 \; \phantom{-}0 \; \phantom{-}0)  \Rightarrow \overline{x_4} = (1 \; -1 \; -2 \; \phantom{-}2)
$$
