How to calculate the inverse matrix by means of elementary matrices?
I know that the method of gauss-jordan is easier
They are essentially the same method! If you can figure out the elementary matrices necessary to reduce matrix A to the identity matrix, then multiplying the identity matrix those elementary matrices, in order, gives $\displaystyle A^{-1}$. How do you find those elementary matrices? By doing the row operations necessary to reduce A to the identity matrix-half of the Gauss-Jordan method. And since multiplying the identity matrix by the elementary matrices is the same as applying those row operations to the identity matrix- the other half of Gauss-Jordan.
For example, to find the inverse of [tex]\left[\begin{array}{cc}1 & 2 \\ 2 & 3\end{array}\right][/itex] by Gause-Jordan
1) Set up the matrix and the identity matrix next to each other:
$\displaystyle \left[\begin{array}{cc}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & 1 \end{array}\right]$
2) Subtract twice the first row from the second:
$\displaystyle \left[\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & 1 \end{array}\right]$
3) Multiply the second row by -1:
$\displaystyle \left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 2 & -1 \end{array}\right]$
4) Subtract twice the second row from the first row:
$\displaystyle \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}-3 & 2 \\ 2 & -1 \end{array}\right]$
The inverse matrix is $\displaystyle \left[\begin{array}{cc}-3 & 2 \\ 2 & -1\end{array}\right]$
Now look at the individual row operations above:
2) Subtract twice the first row from the second. Every row operation corresponds to the elementary matrix you get by performing that row operation on the identity matrix. Subtracting twice the first row of the identity matrix from the second gives $\displaystyle \left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]$.
3) Multiply the second row by -1. Doing that to the identity matrix gives $\displaystyle \left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$.
4) Subract twice the second row from the first. Doing that to the identiy matrix gives $\displaystyle \left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]$.
Doing all those row operations, in order, is the same as multiplying those matrices together.
$\displaystyle \left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]= \left[\begin{array}{cc}-3 & 2 \\ 2 & -1\end{array}\right]$