When trying to find the inverse of the n$\times$n matrix $A$, one way of going about it is by solving $AX=I$, wherein $I$ is the n$\times$n identity matrix, and $X$ is some n$\times$n matrix which is the inverse of $A$. Writing out the matrix product $AX$ will leave you with $n^2$ equations in $n^2$ unknowns. Could someone explain to me how finding the inverse of an invertible matrix $A$ by writing it like this is valid:
$$\left(A|I\right)= \left(
\begin{array}{cccc|cccc}
a_{1,1}&a_{1,2} &\cdots &a_{1,3} &1 &0 &\cdots &0\\
a_{2,1}&a_{2,2} &\cdots &a_{2,3} &0 &1 &\cdots &0\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\
{a_{n,1}}&a_{n,2} &\cdots&a_{n,n} &0 &0 &\cdots &1
\end{array}
\right)$$
It makes sense to do this with a system $\mathit{A}\mathbf{x} = \mathbf{b}$, where $\mathbf{x}, \mathbf{b}$ are column vectors, and $A$ is a coefficient matrix. Solving the augmented matrix shown above isn't difficult; I understand how to do it, and how to get a solution, but I don't understand how it's a valid action to perform. I mean, the identity matrix $I$ to the right isn't a column vector, and as such, when I row-reduce $A$ to the identity matrix, I get:
$$\left(I|C\right)= \left(
\begin{array}{cccc|cccc}
1 &0 &\cdots &0 &c_{1,1} &c_{1,2} &\cdots &c_{1,n}\\
0 &1 &\cdots &0 &c_{2,1} &c_{2,2} &\cdots &c_{2,n}\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\
0 &0 &\cdots &1 &c_{n,1} &c_{n,2} &\cdots &c_{n,n}
\end{array}
\right)$$
which means that each of my variables is equal to a row vector. For example, $\mathbf{x_{1,1}}$ would be $\mathbf{x}_{1,1}=[c_{1,1}, c_{1,2}, \cdots ,c_{1,n}]$. How is this possible? It doesn't make any sense to me at all. It makes me wonder what the variables of the coefficient matrix $A$ are? Apparently, they're row vectors. But how is this even possible, as we were originally trying to solve $AX$: a matrix product which yields only linear equations in the form of dot products of coefficients $a_{ij}$ and variables $x_{ij}$?
Best Answer
Doing row operations is equivalent to multiply your matrix from the left by an elementary matrix, thus you get
$$E_m E_{m-1}\cdot\ldots\cdot E_2E_1A=I$$
Now the above simply means $\;E_m\cdot\ldots\cdot E_1= A^{-1}\;$
You can do the above also with columns operations, which is the same as multiplying your matrix from the right by elementary operations, but never mixed row and column operations: ifyou began with either one, stick to it all the time.