The most common way to find inverse matrix is $M^{-1}=\frac1{\det(M)}\mathrm{adj}(M)$. However it is very trouble to find when the matrix is large.
I found a very interesting way to get inverse matrix and I want to know why it can be done like this. For example if you want to find the inverse of $$M=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$$
First, write an identity matrix on the right hand side and carry out some steps:
$$\begin{bmatrix}1 & 2 &1 &0 \\ 3 & 4&0&1\end{bmatrix}\to\begin{bmatrix}1 & 2 &1 &0 \\ 3/2 & 2&0&1/2\end{bmatrix}\to\begin{bmatrix}1/2 & 0 &-1 &1/2 \\ 3/2 & 2&0&1/2\end{bmatrix}\to\begin{bmatrix}3/2 & 0 &-3 &3/2 \\ 3/2 & 2&0&1/2\end{bmatrix}$$
$$\to\begin{bmatrix}3/2 & 0 &-3 &3/2 \\ 0 & 2&3&-1\end{bmatrix}\to\begin{bmatrix}1 & 0 &-2 &1 \\ 0 & 2&3&-1\end{bmatrix}\to\begin{bmatrix}1 & 0 &-2 &1 \\ 0 & 1&3/2&-1/2\end{bmatrix}$$
You can 1. swap any two row of the matrix 2. multiply a constant in any row 3. add one row to the other row. Just like you are doing Gaussian elimination. when the identical matrix shift to the left, the right hand side become
$$M^{-1}=\begin{bmatrix}-2 &1 \\3/2&-1/2\end{bmatrix}$$
How to prove this method work?
Best Answer
This is a very standard method; if you discovered it on your own, congratulations! It works because each of the elementary row operations that you’re performing is equivalent to multiplication by an elementary matrix. To convert $A$ to $I$, you perform some sequence of elementary row operations, which in effect is multiplying $A$ by a sequence of elementary matrices:
$$I=(E_mE_{m-1}\ldots E_2E_1)A\;,$$
say, if it took $m$ row operations. This says that $E_mE_{m-1}\ldots E_2E_1$ is $A^{-1}$. (Well, actually it says that $E_mE_{m-1}\ldots E_2E_1$ is a left inverse for $A$, but there’s a theorem that says that a left inverse of a square matrix is actually the inverse of the matrix.)
You’ve performed exactly the same operations to $I$ on the other side of the augmented matrix, so on that side you end up with
$$E_mE_{m-1}\ldots E_2E_1I=E_mE_{m-1}\ldots E_2E_1\;,$$
which we just saw is $A^{-1}$.
Thus, if you start with $\begin{bmatrix}A\mid I\end{bmatrix}$, you’re guaranteed to end up with $\begin{bmatrix}I\mid A^{-1}\end{bmatrix}$, exactly as you discovered.