Inverse of a 4x4 Matrix I'm trying to take the inverse of this matrix:
$\left[\begin{array}{rrrr}
2/7 &-6/7  &3/7  &1 \\ 
 6/7& 3/7 &2/7  &2 \\ 
-3/7 &2/7  &6/7  &3 \\ 
0\ \ &0\ \ &0\ \
 &\>\>\>1 
\end{array}\right]$
I have been using this as a guide and I've seen people talk about row elimination (I don't know how to use it) to solve it but I am unsure which strategy would be the best way to attack this problem. I know its going to be a large amount of work either way.
 A: Here the best approach might be to observe that the upper left 3x3 block forms an orthogonal matrix. IOW, those three first columns form an orthonormal set of vectors.
Mind you, that was just a hint. It doesn't give you the inverse of the 4x4 matrix, but it is a good start!
[Edit]
Extending the hint a little bit. The matrix is of the block form
$$
A=\left(\begin{array}{cc}P&X\\0&1\end{array}\right),
$$
where the 3x3 block $P$ is orthogonal, so $P^{-1}=P^T$. Using that observation it is easy to write down an inverse for the matrix $A'$ gotten from $A$ by replacing the the 3-vector $X$ with all zeros. The difference between $A$ and $A'$ amount adding multiples of the fourth row to the others. The way to invert that is...
Looks like the matrix represents a combination of a rotation and a translation of the 3D-space. Did this come from a 3D-graphics programming exercise by any chance?
[/Edit]
A: Here's a method for inverting matrices using row reduction, which I think is what you are after:
Let $A$ be an $n\times n$ matrix. Let $B$ be the $n\times 2n$ matrix obtained by placing the $n\times n$ identity matrix to the right of $A$:
$$B = \left( A\ |\ I_n\right).$$
Now, perform row reduction on this matrix until the left half is the identity matrix (or has a row of zeros). Then the matrix on the right half is $A^{-1}$ (or, if you got a row of zeros in the left half, then $A$ is not invertible).
The reason this works is that performing elementary row operations is equivalent to multiplying on the left by an elementary matrix. By doing the same operations to the identity matrix, you are computing the product of those elementary matrices. If $E_1,\ldots,E_n$ are elementary matrices such that
$$E_nE_{n-1}\cdots E_1A = I_n,$$
then it follows that $E_nE_{n-1}\cdots E_1 = E_{n}E_{n-1}\cdots E_1I_n = A^{-1}$. (Because for $n\times n$ matrices over fields, if $CA=I_n$ then $AC=I_n$). Thus, the computations on the right half of the matrix $B$ give you the inverse of $A$.
For large matrices, this is certainly easier to do by hand than using the adjugate; it is probably faster in computers as well.
A: I think this is better explained in real time.
This video from YouTube seems popular.  It demonstrates the row reduction method to find an inverse on a 3x3.  Try to follow along with the example matrix first, and your matrix will not be much harder.
A: I always find row-reduction much useful and probably fastest for finding inverses.
You can use this dynamic linear algebra tool kit (finding inverse is the fifth feature after you click enter) to learn as well as visualize how your matrix is reduced to its inverse.

ADDED:The direct inverse hyperlink is working  (as of now) and here is the self explained result for your matrix.
