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Step 1
An elementary matrix is a kind of matrices that differ from the identity matrix from a single elementary operation. The elementary operation can be row operation or column operation. Thus after performing one elementary row operation or elementary column operation, the matrix changes to an identity matrix, such type of matrix is called an elementary matrix.
The inverse of elementary matrices is found from the result that when two matrices are multiplied with each other, and the resultant matrix is an identity matrix, then the multiplied matrix is the inverse of the other matrix.
It is represented as,
$A×{A}^{-1}=I$
Here, A is the elementary matrix, ${A}^{-1}$ is the inverse of the elementary matrix and I is the identity matrix.
Step 2
Consider a $\left(3×3\right)$ elementary matrix A shown below: $A=\left[\begin{array}{ccc}1& 0& 0\\ -7& 1& 0\\ 0& 0& 1\end{array}\right]$ The above matrix is a elementary matrix since when a column operation is performed the matrix becomes an identity matrix.
The column operation is $\left({C}_{1}\to {C}_{1}+7×{C}_{2}\right)$, In order to find the inverse of the matrix insert a $\left(3×3\right)$ identity matrix to the left of the elementary matrix and perform the above mentioned column operation to the whole matrix,
Thus, the elementary matrix changes to an identity matrix, and the identity matrix changes to a transformed matrix, and this transformed matrix is the inverse of the elementary matrix.
The steps are shown below:
$A|I=\left[\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ -7& 1& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right]$
$\left({C}_{1}\to {C}_{1}+7×{C}_{2}\right)$
$A|I=\left[\begin{array}{cccccc}1& 0& 0& 1& 0& 0\\ 0& 1& 0& 7& 1& 0\\ 0& 0& 1& 0& 0& 1\end{array}\right]$ Thus, the transformed matrix is,
${A}^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 7& 1& 0\\ 0& 0& 1\end{array}\right]$ This is possible since,
$A×{A}^{-1}=I$
$\left[\begin{array}{ccc}1& 0& 0\\ -7& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 7& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Jeffrey Jordon