Find an elementary matrix such that $EA=B$ I am given two matrices, and I have to find an elementary matrix $A$ such that $EA=B$.
$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$
$$B = \begin{bmatrix}10&4\\-10&-6\end{bmatrix}$$
I tried "transposing" the equation, meaning $(EA)^T = B^T$. The equation given would then be $(A^T)(E^T) = B^T$.
I, however, can't manage to end up with the matrix $B$.
 A: You don't need to go through transposing.
Just find the inverse matrix of $E$
Then $A = E^{-1} B$
How do you find the inverse of $E$? See e.g. here:
https://www.mathsisfun.com/algebra/matrix-inverse.html
See in particular the section 2x2 Matrix
It says that:
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \frac{1}{(ad-bc)} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
OK, so you have
$$E = \begin{bmatrix}2&4\\2&-6\end{bmatrix}$$
Therefore
$$E^{-1} = \frac{1}{(-20)} \begin{bmatrix}-6&-4\\-2&2\end{bmatrix} $$
and then
$$A = \frac{-1}{20} \begin{bmatrix}-6&-4\\-2&2\end{bmatrix} \begin{bmatrix}10&4\\-10&-6\end{bmatrix} =  \begin{bmatrix}1&0\\2&1\end{bmatrix}$$
A: Hint: Can you find a linear combination of the columns of $E$ that would produce the first column of $B$? In other words, can you find constants $k_1$ and $k_2$ such that
$$k_1\begin{bmatrix}2\\2 \end{bmatrix} + k_2\begin{bmatrix}4\\-6 \end{bmatrix} = \begin{bmatrix}10\\-10 \end{bmatrix}?$$
Once you've done that, can you do the same for the second column of $B$? (This will need two more constants, $l_1$ and $l_2$.)
Finally, arrange the four constants you found into a $2 \times 2$ matrix.
