My question is why we have to restrict over selves to find inverse for non singular matrix?
We can define inverse of a square matrix as follows:
A matrix $B$ is said to be inverse of $A$ if $BA = C$, where $C$ is the matrix obtained by $A$ by applying row transformation (some what like normal form). Matrix $C$ must satisfy following properties:
- All zero rows are at the bottom
- leading entry of each non-zero row is $1$.
- $C$ is the matrix obtained by applying row transformation to the maximum extent. The matrix $C$ is identity matrix if $A$ is non singular. Only in that case, $BA = AB = I$ holds.
In this way we can associate with every square matrix a matrix $B$ (inverse) and $C$ (some what normal form of $A$). This will help to solve system of equations even when they have infinite solutions! Am I right?
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$\begingroup$The matrix $B$ you describe represents (by its left multiplication) a combination of row operations that will bring $A$ into into reduced row echelon form (at least I guess that is what you wanted to describe). This is indeed useful for giving the complete solution to linear systems. The main problem with this definition is that if $A$ is singular then $B$ is not unique. Indeed one can left-multiply any such matrix $B$ by any matrix whose $r$ first columns are those of the identity matrix (where $r$ is the rank of $A$), and the other columns are completely arbitrary (if you want $B$ to correspond to a combination of row operations you must ensure that $\det B\neq0$, but that is all, and this still leaves a lot of freedom when $r$ is not maximal).
$\endgroup$ 3 $\begingroup$An inverse of a matrix is one which after matrix multiplication results in an identity matrix (I). So there is no relevance of saying a matrix to be an inverse if it will result in any normal form other than identity.
$\endgroup$ $\begingroup$The Moore-Penrose pseudoinverse is it.
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