Solving the matrix equation and finding the matrices inverse Suppose that $E_1 \begin{bmatrix}12\\35\end{bmatrix} = \begin{bmatrix}48\\35\end{bmatrix}$ Find $E_1$ and $E_1^{-1}$


I set up the equation and saw that $E_1 * A = B$ so I know that:

$E_1 = A^{-1} * B$ 

where $I_m =$ Identity Matrix 

So I attempted to find $A^{-1} = A*I_m = \begin{bmatrix}12&amp;1\\35&amp;0\end{bmatrix}$

After RREF I got: $A^{-1} =\begin{bmatrix}1&amp;1/12\\0&amp;35\end{bmatrix}$

So then I tried $E_1 = \begin{bmatrix}1&amp;1/12\\0&amp;35\end{bmatrix} * \begin{bmatrix}48\\35\end{bmatrix}$ but It is saying that I am incorrect? What am I doing wrong?
Apparently, $E_1$ is a $2\times 2$ matrix. Let, $E = \begin{bmatrix} a&amp;b\\ c&amp;d \end{bmatrix}$, thus we want to solve the linear system:
$$\begin{bmatrix} a&amp;b\\ c&amp;d \end{bmatrix}\cdot \begin{bmatrix}12\\35 \end{bmatrix}=\begin{bmatrix} 48\\35\end{bmatrix}
$$

Thus, we have the system of equations:
$\begin{cases}
12a+35b = 48\\
12c+35d = 35
\end{cases}\implies \begin{cases} b = \frac{48-12a}{35}\\[2ex] d = \frac{35-12c}{35}\end{cases}$

We can choose any values for $a,c$ (as long as $ad-bc\neq 0$ ), otherwise $E^{-1}$ will not be defined. Choosing $a = 1, c = 0\implies b=36/35, d = 1.$ 

So, we have the matrix $E = \begin{bmatrix} 1&amp;\frac{36}{35}\\ 0&amp;1\end{bmatrix}.$ You can indeed confirm that $\begin{bmatrix} 1&amp;\frac{36}{35}\\ 0&amp;1\end{bmatrix}\cdot \begin{bmatrix} 12\\35 \end{bmatrix} = \begin{bmatrix} 48\\35 \end{bmatrix}.$

Now, since $E$ is $2\times 2$, we can easily find the inverse. If $E = \begin{bmatrix} a&amp;b\\c&amp;d\end{bmatrix}$, then the inverse is given by the formula:
$$E^{-1} = \frac{1}{ad-bc}\cdot \begin{bmatrix} d&amp;-b\\-c&amp;a\end{bmatrix}.$$
there are errors:

even in the case defined $EA = B$ does not imply $E = A^{-1}B$,
for  a matrix A , 

the inverse matrix of $A$ (denoted $A^{-1}$)
exists iff the matrix $A$ is a square matrix and $detA \neg 0$.

you say " so I attempted  to find $A^{-1}=A*I_n$" there is no
reason in the text to look for this type of matrix