My professor gave us this formula for finding the inverse of 3x3 matrices;
[matrix given|identity] then row reduce matrix given to the identity and you end up with [identity|inverse of matrix given]
So basically you take the matrix you are trying the find the inverse of, augment it with the identity, row reduce it until you have the identity for the first 3 columns, and then your last 3 columns will be the inverse. (I hope my wording makes sense.)
For example:
to find the inverse of \begin{pmatrix}0&1&2\\ 1&0&3\\ 4&-3&8\end{pmatrix}
you first do this: \begin{pmatrix}0&1&2&1&0&0\\ 1&0&3&0&1&0\\ 4&-3&8&0&0&1\end{pmatrix} then row reduce it until you have the identity at the beginning and the last 3 columns will be the inverse:
\begin{pmatrix}1&0&0&-\frac{9}{2}&7&-\frac{3}{2}\\ 0&1&0&-2&4&-1\\ 0&0&1&-\frac{3}{2}&-2&\frac{1}{2}\end{pmatrix}
so the inverse is: \begin{pmatrix}-\frac{9}{2}&7&-\frac{3}{2}\\ -2&4&-1\\ \frac{3}{2}&-2&\frac{1}{2}\end{pmatrix}
My question is: does this work for other matrices bigger than a 3x3? Thanks!
The answer is yes. The reason is, that performing row operations, and row reducing, are instances of multiplication on the left by a square matrix. If your starting big matrix is $(A,I)$ and your ending matrix is $(I,B)$, you got there by multiplying by some matrix $M$ on the left, with $M \times (A,I) = (I,B)$. That is, $M\times A=I$ and $M\times I=B$. Clearly $M=B$, and since $MA=I$, we have $M=A^{-1}.$ Here $M$ is the combined effect of all the row operations you performed in reducing $A$; the matrix $B$ in effect "bookkeeps" what you did.