How to get inverse of this matrix using least amount of space? I'm working on a problem from a past exam and I'm stuck, so I'm asking for help.
Here it is:  $A = \frac12 \left[\begin{array}{rrrr}
1 &  1 &  1 &  1 \\
1 &  1 & -1 & -1 \\
1 & -1 &  1 & -1 \\
1 & -1 & -1 &  1
\end{array}\right]$ find  $\mathbf A^{-1}$.
My problem isn't the inverse matrix itself. We just get the determinant, see if it's zero or not, get the adjoint matrix and divide it by determinant.   
My problem is space. As you can see, it's a 4x4 matrix meaning that I'd have to do 4x4 3x3 determinants to get the adjoint matrix plus 2 3x3 determinants to get determinant of the matrix. Now we get one A3 piece of paper for 6 problems. The problems are printed on one side and the other side is blank. This and the fact that inverse matrix is $A = \frac12 \left[\begin{array}{rrrr}
1 &  1 &  1 &  1 \\
1 &  1 & -1 & -1 \\
1 & -1 &  1 & -1 \\
1 & -1 & -1 &  1
\end{array}\right]$
led me to believe that there's some catch that I do not see. Any ideas what could it be?
Also if someone could edit these matrices from MATLAB format into something that this site will parse would be great!
EDIT Unfortunately it seem that TeX code for matrices doesn't work here. Here's the matrix in MATLAB form, if anyone wants it A=(1/2)*[1,1,1,1;1,1,-1,-1;1,-1,1,-1;1,-1,-1,1];
EDIT 2 Answer by Jack Schmidt contains code for matrices. 

Gauss/Jordan Elimination will do it.  It'll let you find |A|^1 with out the bother of finding the determinant. Just augment your original matrix with the identity and let her rip.
On an aside, you can still deduce the determinant from the inverse.
{ |A|^1= (1/det)[adj|A|]
therefore the determinant is equal to the lowest common denominator of all of the elements of the inverse.
